Logarithmic Functions

The section covers:

Half-Life problems can be found here.

Introduction to Logarithms (Logs)

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What is a Log and Why do we Need Them?

I have to admit that logs are one of my favorite topics in math. I’m not sure exactly why, but you can do so many awesome things with them!

We’ll soon see that Logs can be used to “get the variable in the exponent down” so we can solve for it. But logarithms are also used for many other things, including early on to perform computations – before calculators and computers were around. Have you ever heard of a slide rule? (Ask your parents…or grandparents…)

A slide rule was used (among other things) to multiply and divide large numbers by adding and subtracting their exponents. The numbers on the slide rules had different scales (“logarithmic scales”, meaning that the distance between numbers increase exponentially) and you could simply look up a number, and slide the ruler over to another number to get the number you want. When doing this, you were adding and subtracting exponents, thus multiplying and dividing large numbers. Genius!

Definition of Logarithm

Remember:  A log is in exponent! So when you take the log of something, you are getting back an exponent. The two equations below are two different ways to say the same thing, but the first is an exponential equation, and the second is a logarithmic equation.

Note that \(b\) is called the base of the log, and must be greater than 0 (so we don’t have to deal with complex numbers). Also, the base can’t be 1, or the equations wouldn’t be exponential or logarithmic.

The \(x\) in the log equation is called the argument and it must be greater than 0, again, to avoid complex numbers.

Exponential Function                         Logarithmic Function

\(x={{b}^{y}}\)                               \(\Leftrightarrow \)                              \(y={{\log }_{b}}x\)

\((b>0,\,\,b\ne 1,\,\,x>0)\)

Example:                                                           Example:

   \(\displaystyle 25={{5}^{2}}\)                                                            \(\displaystyle 2={{\log }_{5}}25\)

To illustrate how these two equations are related, many times a “loop” is shown, which shows that \(b\) raised to the \(y\) equals \(x\). Again, \(x\) is called the argument of the log, and you can write it as \({{\log }_{b}}x\) or \({{\log }_{b}}(x)\). Learn this well!

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Note: If there is no \(b\) next to the log, then the base is assumed to be 10.

Here are some examples where we change an exponential function to a log function, and a log function to an exponential function. See the loop?

Exponential Form

Log Form

 \({{2}^{3}}=8\) \({{\log }_{2}}8=3\)
 \(\displaystyle {{\left( 2 \right)}^{{-4}}}=\frac{1}{{16}}\)  \(\displaystyle {{\log }_{2}}\frac{1}{{16}}=-4\)
 \({{4}^{y}}=x\)  \({{\log }_{4}}x=y\)

Here are some simple log problems where we have to use what we know about exponents to find the log back. You’ll probably have some of these to work on tests without a calculator:

Log Problem

How to Solve:    The Log IS the Exponent!

 \({{\log }_{4}}4=x\) 4 raised to ? = 4. Since \({{4}^{1}}=1,\,\,x=1\).
 \({{\log }_{3}}81=x\) 3 raised to ? = 81. Since \({{3}^{4}}=81,\,\,x=4\).
 \(\displaystyle {{\log }_{2}}\frac{1}{{16}}=x\) 2 raised to ? \(\displaystyle =\frac{1}{{16}}\). Since \(\displaystyle {{2}^{4}}=16,\,\,\,\,{{2}^{{-4}}}=\frac{1}{{16}}\), so \(x=-4\).
 \(\displaystyle {{\log }_{{32}}}\frac{1}{4}=x\) 32 raised to ? = \(\displaystyle \frac{1}{4}\). Since this is a little tricky to figure out with mental math, we can use the “solving exponents by matching bases” method:

\(\displaystyle {{32}^{x}}=\frac{1}{4},\,\,\,\,\,{{\left( {{{2}^{5}}} \right)}^{x}}={{2}^{{-2}}},\,\,\,\,\,\,{{2}^{{5x}}}={{2}^{{-2}}},\,\,\,\,5x=-2,\,\,x=-\frac{2}{5}\).

 \({{\log }_{4}}32=x\) 4 raised to ? = 32.  We know that \({{2}^{5}}=32\), and we also know that \(\sqrt{4}={{4}^{{\frac{1}{2}}}}=2\). Remember that the root is on the bottom of an exponential fraction, and the “raising to the power” is on the top, which would make \({{4}^{{\frac{5}{2}}}}=32\) (we can also multiply 5 by \(\displaystyle \frac{1}{2}\)). So \(\displaystyle x=\frac{5}{2}\).

We could also solve this using the matching base method, as we did in the previous problem.

 \({{\log }_{b}}1=x\) \(b\) (any number) raised to ? = 1. Since \({{b}^{0}}=1,\,\,x=0\).

Special Logarithms

Most of the logarithms that you’ll work with have either a base of “10” (because we’ll deal in base 10 with our counting system) or base “\(e\)”.

A logarithm with base 10 is called a common logarithm, and when you see “log” without a small subscript for the base, you assume it is base 10. Thus, \(\log \left( 1000 \right)=3\)  and  \({{\log }_{10}}\left( 1000 \right)=3\), but we don’t need the 10.

A logarithm with base \(e\) is called the “natural logarithm” and is written as \(\ln \left( x \right)\). Thus, we write  \({{\log }_{e}}\left( x \right)\) as \(\ln \left( x \right)\). Again, the base “\(e\)” has many applications in both engineering and economics.

Using Logs (and Exponents) in the Graphing Calculator

You can use graphing calculator keys to find the basic logs: LOG (base 10) and LN.  For logs with other bases, you can use a function called LOGBASE under MATH (or ALPHA WINDOW 5), or use what we call a “change of base” formula, that we’ll introduce here and talk about more in Basic Log Properties below.

We learned how to put exponents in the calculator (using ^)  here in the Exponents and Radicals in Algebra section.


Note that you can also use your calculator to perform logarithmic regressions, using a set of points, like we did here in the Exponential Functions section.

Parent Graphs of Logarithmic Functions

Here are some examples of parent log graphs. I always remember that the “reference point” (or “anchor point“) of a log function is \((1,0)\) (since this looks like  the “lo” in “log”). We have to also remember that if the function shifts, this “anchor point” will move. The graph of an log function (a parent function: one that isn’t shifted) has an asymptote of \(x=0\).

When the base is greater than 1 (a growth), the graph increases, and when the base is less than 1 (a decay), the graph decreases. The domain and range are the same for both parent functions.

Remember from Parent Graphs and Transformations that the critical or significant points of the parent logarithmic function \(y={{\log }_{b}}x\) (inverse of exponential function) are \(\displaystyle \left( {\frac{1}{b},\,-1} \right),\,\left( {1,0} \right),\,\left( {b,1} \right)\).

“Parent” Log Graphs

Domain:  \(\left( {0,\infty } \right)\);   Range: \(\left( {-\infty ,\infty } \right)\)

 

End Behavior: \(\left\{ \begin{array}{l}x\to {{0}^{+}},\,\,y\to -\infty \\x\to \infty ,\,\,\,y\to \infty \end{array} \right.\)

Domain:  \(\left( {0,\infty } \right)\);   Range:   \(\left( {-\infty ,\infty } \right)\)

 

End Behavior:  \(\left\{ \begin{array}{l}x\to {{0}^{+}},\,\,y\to -\infty \\x\to \infty ,\,\,\,y\to \infty \end{array} \right.\)

Transformations of Log Functions

Remember again that the generic equation for a transformation with vertical stretch \(a\), horizontal shift \(h\), and vertical shift \(k\) is \(f\left( x \right)=a\cdot \log \left( {x-h} \right)+k\) for log functions.

Remember these rules:
When functions are transformed on the outside of the \(f(x)\) part, you move the function up and down and do “regular” math, as we’ll see in the examples below. These are vertical transformations or translations.

When transformations are made on the inside of the \(f(x)\) part, you move the function back and forth (but do the opposite math – basically since if you were to isolate the x, you’d move everything to the other side). These are horizontal transformations or translations.

When there is a negative sign outside the parentheses, the function is reflected (flipped) across the \(x\)-axis; when there is a negative sign inside the parentheses, the function is reflected across the \(y\)-axis.

For log functions, get the new asymptote by setting the log argument to 0 and solving for \(x\) (horizontal shift). The domain changes with the horizontal shift, and the range is always \(\left( {-\infty ,\infty } \right)\).

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Here are some examples, using t-charts:

Transformation

T-chart

Graph

\(\begin{array}{c}y=\log \left( {2x-2} \right)-1\\y=\log \left( {2\left( {x-1} \right)} \right)-1\end{array}\)

 

Parent function:

\(y=\log \left( x \right)\)

 

For log and ln functions, use –1, 0, and 1 for the y values for the parent function.

To get new asymptote, set the log argument to 0 and solve for x.

\(\frac{1}{2}x+1\) x   y y – 1
 1.05 .1 –1 –2
1.5 1  0 –1
6 10  1   0

 Asymptote:  \(x=1\)

Domain:  \(\left( {1,\,\,\infty } \right)\)

Range:  \(\left( {-\infty ,\,\,\infty } \right)\)

Horizontal compression by a factor of \(\frac{1}{2}\), translate right 1 unit, down 1 unit.

\(y=\ln \left( {-x} \right)+5\)

 

Parent function:

\(y=\ln \left( x \right)\)

 

A negative sign inside the parentheses means reflecting across the y-axis.

Try without a T-chart.

Reflect function across the y-axis, so it ends up going down.

Function is also shifted up 5, since instead of going through \(\left( {-1,0} \right)\), it goes through \(\left( {-1,5} \right)\).

Asymptote:  \(x=0\)

Domain:  \(\left( {-\infty ,0} \right)\)

Range:  \(\left( {-\infty ,\,\,\infty } \right)\)

Reflect over the y-axis, translate up 5 units.

Writing Logarithmic Equations from Points and Graphs

You may be also be asked to write log equations, such as the following:

  • Write an equation to describe the logarithmic function in form \(y=a{{\log }_{b}}x\), with a given base and a given point.
  • Write a logarithmic function in form \(y=a\log \left( {x-h} \right)+k\) from a graph (given asymptote and two points).

Note: we may also be able to use logarithmic regression to find logs equations based on points, like we did here with Exponential Regression.

Let’s try these types of problems:

Problem Solution
Write an equation to describe the logarithmic function in form \(y=a{{\log }_{b}}x\), with base 3 and passing through the point \(\left( {81,\,2} \right)\). The equation will be in the form \(y=a{{\log }_{3}}x\), since the base is 3. Plug in 81 for \(x\) and 2 for \(y\), and solve for \(a\):

\(\begin{array}{l}y=a{{\log }_{3}}x\\2=a{{\log }_{3}}81=a\cdot 4\,\,\,({{3}^{4}}=81)\\a=\frac{2}{4}=\frac{1}{2}\end{array}\)

 

The equation is \(y=.5{{\log }_{3}}x\).

For the following graph, write the appropriate logarithmic function in the form \(y=a\log \left( {x-h} \right)+k\):

 

 

We see that this graph has an asymptote at \(x=3\), so it will have a horizontal shift of 3, or \(h=3\). Our equation will be in the form \(y=a\log \left( {x-3} \right)+k\).

 

Now we need to use a system of equations with the two points on the graph: \(\left( {4,2} \right)\) and \(\left( {13,0} \right)\):

 

We can solve for \(a\) first using \(\left( {4,2} \right)\):

 

\(\displaystyle \begin{array}{l}y=a\log \left( {x-3} \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=a\log \left( {x-3} \right)+2\\2=a\log \left( {4-3} \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=a\log \left( {13-3} \right)+2\\2=a\log \left( 1 \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=a\log \left( {10} \right)+2\,\,\,\,\\2=a\cdot 0+k;\,\,\,\,\,k=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=a\cdot 1+2;\,\,\,\,\,a=-2\end{array}\)

 

The logarithmic function is \(y=-2\log \left( {x-3} \right)+2\). Graph this function and it works!   √

 

Basic Log Properties, Including Shortcuts

When working with logs, there are certain shortcuts that you can use over and over again. It’s important to understand these, but later, when using them, be familiar with them, so you can use them quickly. Think of some of the rules as “canceling out” the logs with the exponents. And don’t forget the log “loop”:

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Log Rule

Explanation

Example

 \({{\log }_{a}}a=1\)

Loop: \(a\) raised to 1 is \(a\).

 \(\log 10=1\)
 \({{\log }_{a}}1=0\)

Loop: \(a\) raised to 0 is 1.

 \({{\log }_{2}}1=0\)
 \(\ln e\,\,\left( {={{{\log }}_{e}}\left( e \right)} \right)\,\,=1\)

Loop: \(e\) raised to 1 is \(e\).

 \(\ln e=1\)
 \({{\log }_{a}}{{a}^{x}}=x\)

Loop: \({{a}^{x}}={{a}^{x}}\).

 \({{\log }_{2}}{{2}^{4}}=4\)
 \(\ln {{e}^{x}}\,\,\left( {={{{\log }}_{e}}\left( {{{e}^{x}}} \right)} \right)=x\)

Loop: \({{e}^{x}}={{e}^{x}}\).

 \(\ln {{e}^{2}}=2\)
 

\({{a}^{{{{{\log }}_{a}}x}}}=x\)

\(a\) raised to “the exponent that a is raised to get \(x\)” is \(x\).

(This is a difficult concept; you have to really think about it!)

 \({{4}^{{{{{\log }}_{4}}\left( 3 \right)}}}=3\)
 \({{e}^{{\ln x}}}=x\)

\(e\) raised to “the exponent that \(e\) is raised to get \(x\)” is \(x\).

 \({{e}^{{\ln 5}}}=5\)

Now to the important log properties!  These properties  are derived from the fact that we add exponents when we multiply terms with exponents, we subtract exponents when we divide, and we multiply exponents when we raise them to a power. These are powerful properties that we’ll need to use to isolate variables in exponents so we can solve for them. You WILL have to memorize these and remember that for the first three, you must be dealing with logs with the same base:

Summary of Logarithmic Properties

(for all formulas, \(b,x,y>0,\,\,\,b\ne 1\))

Product Rule:

 \({{\log }_{b}}\left( {xy} \right)={{\log }_{b}}x+{{\log }_{b}}y\) Remember, if you multiply inside the parentheses, you add logs on the outside.

Division Rule:

 \(\displaystyle {{\log }_{b}}\left( {\frac{x}{y}} \right)={{\log }_{b}}x-{{\log }_{b}}y\) Remember, if you divide inside the parentheses, you subtract logs on the outside.

Power Rule:

 \({{\log }_{b}}\left( {{{x}^{p}}} \right)=p{{\log }_{b}}x\) Remember, if you raise to a power inside the parentheses, you multiply by that power on the outside.

Change of Base Rule:

 \(\displaystyle {{\log }_{b}}\left( x \right)=\frac{{{{{\log }}_{{10}}}x}}{{{{{\log }}_{{10}}}b}}\)  or  \(\displaystyle {{\log }_{b}}\left( x \right)=\frac{{\ln x}}{{\ln b}}\) Remember, put the larger number (argument) on top, the smaller (base) on bottom. This is useful for the calculator if you don’t have the LOGBASE function.

And don’t forget the basics:

\(y={{\log }_{b}}x\,\,\,\,\,\Leftrightarrow \,\,\,\,\,\,x={{b}^{y}}\,\,\,\)  \({{b}^{x}}={{b}^{y}}\,\,\,\,\Leftrightarrow \,\,\,\,x=y\)  \({{\log }_{b}}x={{\log }_{b}}y\,\,\,\,\,\Leftrightarrow \,\,\,\,\,x=y\)
\(\left( {b,x,y>0,\,\,\,b\ne 1} \right)\)

Expanding and Condensing Logs

Now we’re going to use these properties to expand and condense logs. Expanding Logs generally means turning the “inside multiplying” (with only one log) to “outside adding” (with multiple logs). Condensing Logs generally means turning the “outside adding” (with multiple logs) to the “inside multiplying” (with only one log).

Why do we need to do all this? We will need to expand and condense logs to solve log problems.

Note that when expanding logs, it’s generally a good idea to apply the power rule last (unless whole terms are raised to a power, as in the third example). Also, note that these logs can be written with or without the arguments in parentheses (for example, as \({{\log }_{3}}5{{x}^{3}}\,\,\,\text{or}\,\,\,\,{{\log }_{3}}\left( 5{{x}^{3}} \right)\), which is different than \({{\left( {{\log }_{3}}5x \right)}^{3}}\)).

Single Log Expanded Log Notes
 \({{\log }_{3}}5{{x}^{3}}\)  \({{\log }_{3}}5+{{\log }_{3}}{{x}^{3}}={{\log }_{3}}5+3{{\log }_{3}}x\)

First, separate the factors with two different logs; then, move the 3 down and around to the front.

 \(\displaystyle {{\log }_{2}}\frac{{2x}}{y}\)  \({{\log }_{2}}2+{{\log }_{2}}x-{{\log }_{2}}y=1+{{\log }_{2}}x-{{\log }_{2}}y\)

Everything on top has a plus before it, and everything on bottom has a minus before it. (The \({{\log }_{2}}2\) has an invisible plus before it). Then, simplify the \({{\log }_{2}}2\) to 1. It’s almost like the 2’s cancel out.

 \({{\log }_{4}}{{\left( {5x} \right)}^{4}}\) \(4{{\log }_{4}}\left( {5x} \right)=4\left( {{{{\log }}_{4}}5+{{{\log }}_{4}}x} \right)=4{{\log }_{4}}5+4{{\log }_{4}}x\)

or

\({{\log }_{4}}{{5}^{4}}{{x}^{4}}={{\log }_{4}}{{5}^{4}}+{{\log }_{4}}{{x}^{4}}=4{{\log }_{4}}5+4{{\log }_{4}}x\)

Since the whole term is raised to 4, first move the 4 down around to the front. Then separate the factors, and finally push through the 4. You can also “distribute” the exponent 4 to the 5 and \(x\), and then “pull things apart”.

 \(\displaystyle \ln \frac{e}{{4{{x}^{3}}}}\)  \(\ln e-\ln 4-\ln {{x}^{3}}=1-\ln 4-3\ln x\)

Remember again that everything on top has a plus before it, and everything on bottom has a minus before it. Simplify the \(\ln e\) to 1, and move the 3 around to the front.

 \(\displaystyle {{\log }_{2}}\sqrt[4]{{\frac{{5x}}{{4{{y}^{2}}}}}}\)  \(\displaystyle \begin{align}\frac{1}{4}{{\log }_{2}}\frac{{5x}}{{4{{y}^{2}}}}&=\frac{1}{4}\left( {{{{\log }}_{2}}5+{{{\log }}_{2}}x-{{{\log }}_{2}}4-2{{{\log }}_{2}}y} \right)\\&=\frac{1}{4}\left( {{{{\log }}_{2}}5+{{{\log }}_{2}}x-2-2{{{\log }}_{2}}y} \right)\\&=\frac{1}{4}{{\log }_{2}}5+\frac{1}{4}{{\log }_{2}}x-\frac{1}{2}-\frac{1}{2}{{\log }_{2}}y\end{align}\)

Since everything is under the root, first move the  around to the front (the 4th root means raised to the \(\displaystyle \frac{1}{4}\)). Keep the \(\displaystyle \frac{1}{4}\) out in front, but expand the fraction; again, everything on top has a plus before it, and everything on bottom has a minus. Then push through the \(\displaystyle \frac{1}{4}\). We also needed to simplify the \({{\log }_{2}}4\) to 2 \(({{2}^{2}}=4)\).

Now let’s go the other way. When condensing logs, it’s generally better to apply the power rule first.

Expanded Log Condensed or Single Log Notes
 \({{\log }_{3}}8+{{\log }_{3}}x-{{\log }_{3}}y\)  \(\displaystyle {{\log }_{3}}\frac{{8x}}{y}\) Everything with a \(+\) sign before it goes on top (numerator), and with a – sign before it goes on bottom (denominator). Remember that there is an invisible \(+\) before the first term.
 \(\displaystyle 4{{\log }_{2}}4+\frac{1}{2}{{\log }_{2}}x\) \(\displaystyle \begin{align}{{\log }_{2}}{{4}^{4}}+{{\log }_{2}}{{x}^{{\frac{1}{2}}}}&=\,{{\log }_{2}}\left( {{{4}^{4}}{{x}^{{\frac{1}{2}}}}} \right)\,\,\,\\\,\,\,\,\,\,\,&=\,{{\log }_{2}}\left( {256\sqrt{x}} \right)\end{align}\) Apply the power rule first by moving up the coefficients and making them exponents. Then change the \(+\) on the outside to a “times” on the inside. Note that we could have simplified the \({{\log }_{2}}{{4}^{4}}\) to 8 before or after we condensed the logs. Also note that \({{\log }_{2}}{{4}^{4}}\) (8) is not the same as \({{\left( {{{{\log }}_{2}}4} \right)}^{4}}\) (16).
 \(\displaystyle \ln x+5\ln \left( {5y} \right)-\frac{1}{3}\ln z\) \(\displaystyle \begin{align}\ln x+\ln {{\left( {5y} \right)}^{5}}-\ln {{z}^{{\frac{1}{3}}}}\,&=\,\ln \left[ {\frac{{x{{{\left( {5y} \right)}}^{5}}}}{{\sqrt[3]{{\ln z}}}}} \right]\\\,\,\,\,\,\,&=\ln \left( {\frac{{3125x{{y}^{5}}}}{{\sqrt[3]{{\ln z}}}}} \right)\end{align}\) Apply the power rule first (moving the coefficients up to be exponents), and then condense the logs by moving everything with a \(+\) sign before it on the top and everything with a – before it sign on the bottom. We can then take the 5th power of 5.
 \(\begin{array}{l}{{\log }_{2}}\left( {{{x}^{2}}-2x-3} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,-{{\log }_{2}}\left( {x-3} \right)\end{array}\) \(\displaystyle \begin{align}{{\log }_{2}}\frac{{\left( {{{x}^{2}}-2x-3} \right)}}{{\left( {x-3} \right)}}\,&=\,{{\log }_{2}}\frac{{\left( {x-3} \right)\left( {x+1} \right)}}{{\left( {x-3} \right)}}\\&={{\log }_{2}}\left( {x+1} \right)\end{align}\) Condense the log by moving the \({{x}^{2}}-2x-3\) to the top and the \(x-3\) to the bottom.Then factor, and reduce. Note in this problem that \(x\ne 3\).
 \(\displaystyle \frac{1}{3}\left[ {4\ln \left( {x+2} \right)-\ln x-\ln y} \right]\) \(\displaystyle \frac{1}{3}\ln \left[ {\frac{{{{{\left( {x+2} \right)}}^{4}}}}{{xy}}} \right]\,=\,\ln \sqrt[3]{{\left( {\frac{{{{{\left( {x+2} \right)}}^{4}}}}{{xy}}} \right)}}\) Since the \(\frac{1}{3}\) is on the outside, apply that last. First condense the logs inside the [ ] by moving everything with a \(+\) sign before it on the top and everything with a – sign before it on the bottom. Then, because of the \(\frac{1}{3}\), take the cube root of the argument.
\(\displaystyle \begin{array}{l}\log x+\frac{1}{2}\log \left( {1-x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4\log \left( {x+1} \right)\end{array}\)  \(\displaystyle \begin{array}{l}\log x+\log {{\left( {1-x} \right)}^{{\frac{1}{2}}}}-\log {{\left( {x+1} \right)}^{4}}\,\,\,\\\,\,\,\,\,=\,\,\,\log \left[ {\frac{{x\sqrt{{1-x}}}}{{{{{\left( {x+1} \right)}}^{4}}}}} \right]\end{array}\) Move the coefficients up to exponents first, and then condense the logs by moving everything with a \(+\) sign before it on the top and everything with a – sign on the bottom.
 \(\displaystyle \begin{array}{l}{{\log }_{6}}{{6}^{{12x}}}-{{\log }_{6}}{{216}^{x}}\\\,\,\,\,\,\,\,\,\,+2{{\log }_{6}}{{\left( {\frac{1}{{36}}} \right)}^{x}}\end{array}\)  \(\require{cancel} \begin{align}{{\cancel{{{{{\log }}_{6}}6}}}^{{12x}}}&-x{{\log }_{6}}216+2{{\log }_{6}}{{\left( {{{6}^{{-2}}}} \right)}^{x}}\\&=12x-x\cdot 3+{{\log }_{6}}{{\left( {{{6}^{{-2x}}}} \right)}^{2}}\\&=12x-3x+{{\cancel{{{{{\log }}_{6}}6}}}^{{-4x}}}\\&=9x-4x=5x\end{align}\) This one’s a little different, since we can first simplify each of the terms instead of creating one log (which would have worked, too!).

Remember that when when we raise an exponent to another exponent, we multiply exponents.

 

Solving Exponential Equations using Logs

Now we can use all these tools to solve log equations! Remember again that math is just using tools that you have to learn to solve problems.

Remember to always check your answer to make sure the argument of logs (what’s directly following the log) is positive!

Let’s just jump in and trying some solving. There are the basic ways to solve log problems:

1.  Use the power rule to get the exponent down if the variable is in the exponent (probably the most commonly used “tool”). Before doing this, get the base/exponent by itself and take the ln or log of each side. You’ll typically use this when you don’t have logs in the equation, but have variables in the exponents. We typically use ln instead of log, unless we’re dealing with base 10.

We can also use the log with the base under the exponent, as in the second case of the first example:

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Solving Logs:  Use the Power Rule to Bring Exponent Down

\(\displaystyle 150{{\left( {.5} \right)}^{t}}=1500\)

 

\(\displaystyle \begin{align}{{\left( {.5} \right)}^{t}}&=\frac{{1500}}{{150}}=10\\\ln {{\left( {.5} \right)}^{t}}&=\ln \left( {10} \right)\\t\ln \left( {.5} \right)&=\ln \left( {10} \right)\\t&=\frac{{\ln \left( {10} \right)}}{{\ln \left( {.5} \right)}}\approx -3.32\end{align}\)or    \(\displaystyle \begin{align}{{\left( {.5} \right)}^{t}}&=\frac{{1500}}{{150}}=10\\{{\log }_{{.5}}}{{\left( {.5} \right)}^{t}}&={{\log }_{{.5}}}\left( {10} \right)\\t&={{\log }_{{.5}}}\left( {10} \right)\approx -3.32\end{align}\)

\(\displaystyle {{5}^{{x-2}}}=1000\)

 

\(\displaystyle \begin{align}\log {{5}^{{x-2}}}&=\log 1000\\\left( {x-2} \right)\log 5&=3\\x-2&=\frac{3}{{\log 5}}\\x&=\frac{3}{{\log 5}}+2\approx 6.29\end{align}\)

\(\displaystyle {{2}^{{x+3}}}={{5}^{{3x}}}\)

 

\(\displaystyle \begin{align}\ln {{2}^{{x+3}}}&=\ln {{5}^{{3x}}}\\\left( {x+3} \right)\ln 2&=\left( {3x} \right)\ln 5\\x\ln 2+3\ln 2&=3x\ln 5\\3x\ln 5-x\ln 2&=3\ln 2\\x\left( {3\ln 5-\ln 2} \right)&=3\ln 2\\x&=\frac{{3\ln 2}}{{3\ln 5-\ln 2}}\approx .503\end{align}\)

Note: You might be asked to solve simple problems like these using the log “loop” and change of base formula instead:

Solving Logs: Use the Log “Loop” and Change of Base to get Exponent Down

 \(\displaystyle 150{{\left( {.5} \right)}^{t}}=1500\)

 

\(\displaystyle \begin{align}{{\left( {.5} \right)}^{t}}&=\frac{{1500}}{{150}}=10\\{{\log }_{{.5}}}\left( {10} \right)&=t\\t&={{\log }_{{.5}}}\left( {10} \right)\\t&=\frac{{\log \left( {10} \right)}}{{\log \left( {.5} \right)}}\approx -3.32\end{align}\)

 \({{5}^{{x-2}}}=1000\)

 

\(\begin{align}{{\log }_{5}}\left( {1000} \right)&=x-2\\x&={{\log }_{5}}\left( {1000} \right)+2\\x&=\frac{{\log \left( {1000} \right)}}{{\log \left( 5 \right)}}+2\approx 6.29\end{align}\)

2.  If the same log and same base are on both sides, you can just set arguments of logs equal to each other; this is called the “One-to-One Property of Logarithms”:

Solving Logs: Set Arguments of Logs with Same Base Equal to Each Other

 \(3\ln 4=\ln x\)

\(\begin{align}\ln {{4}^{3}}&=\ln x\\{{4}^{3}}&=x\\x&=64\end{align}\)

 \({{\log }_{6}}\left( {2x-5} \right)+{{\log }_{6}}x={{\log }_{6}}10\)

 

\(\begin{align}{{\log }_{6}}\left[ {\left( {2x-5} \right)\left( x \right)} \right]&={{\log }_{6}}10\\2{{x}^{2}}-5x&=10\\2{{x}^{2}}-5x-10&=0\\x&=\frac{{5\pm \sqrt{{25-\left( 4 \right)\left( 2 \right)\left( {-10} \right)}}}}{4}=\frac{{5\pm \sqrt{{105}}}}{4}\\x&=\frac{{5+\sqrt{{105}}}}{4}\text{ }\,\,\,\text{ (negative root won }\!\!’\!\!\text{ t work)}\end{align}\)

3.  If \(x\) is underneath a complicated exponent, raise each side to the reciprocal of that exponent. You typically use this if you have variables raised to exponents.  These are not really log problems, but you’ll see them:

Solving Logs: Raise Each Side to Reciprocal of Exponent

 \(5{{x}^{{\frac{2}{3}}}}=125\)

 

\(\require{cancel} \begin{align}{{x}^{{\frac{2}{3}}}}&=25\\{{\left( {{{x}^{{\cancel{{\frac{2}{3}}}}}}} \right)}^{{\cancel{{\frac{3}{2}}}}}}&={{25}^{{\frac{3}{2}}}}\\x&=125\end{align}\)

 \(3+{{(x+2)}^{{\frac{1}{4}}}}=5\)

 

\(\begin{align}{{(x+2)}^{{\frac{1}{4}}}}&=2\\{{\left( {{{{(x+2)}}^{{\cancel{{\frac{1}{4}}}}}}} \right)}^{{\cancel{{\frac{4}{1}}}}}}&={{2}^{4}}\\x+2&=16\\x&=14\end{align}\)

 \({{x}^{{2.5}}}=32\)

 

\(\begin{align}{{x}^{{\frac{5}{2}}}}&=32\\{{\left( {x\cancel{{^{{\frac{5}{2}}}}}} \right)}^{{\cancel{{\frac{2}{5}}}}}}&={{32}^{{\frac{2}{5}}}}\\x&=4\end{align}\)

4.  Use the loop to get the variable out of the log argument. You typically use this if you have logs in the equations. Remember to get the log by itself before you use the “loop”.

Solving Logs: Use the Log “Loop”

 \(\displaystyle {{\log }_{8}}x=-\frac{1}{3}\)

 

 

\(\displaystyle \begin{align}{{8}^{{-\frac{1}{3}}}}=x\\x=\frac{1}{{\sqrt[3]{8}}}=\frac{1}{2}\end{align}\)

\(\ln \left( {x+3} \right)+\ln \left( {x-3} \right)=4\)

 

\(\begin{align}\ln \left( {\left( {x+3} \right)\left( {x-3} \right)} \right)&=4\\\ln \left( {{{x}^{2}}-9} \right)&=4\\{{\log }_{e}}\left( {{{x}^{2}}-9} \right)&=4\\{{x}^{2}}-9&={{e}^{4}}\\{{x}^{2}}&={{e}^{4}}+9\approx 63.598\\x&\approx \pm \sqrt{{63.598}}\approx 7.97\end{align}\)

 

(Negative root won’t work)

 \({{\log }_{5}}\left( {{{x}^{2}}+4x+4} \right)=2\)

 

\(\begin{align}{{x}^{2}}+4x+4&={{5}^{2}}\\{{x}^{2}}+4x+4-25&=0\\{{x}^{2}}+4x-21&=0\\\left( {x+7} \right)\left( {x-3} \right)&=0\\x=-7,\,\,\,\,x=3\end{align}\)

 

Both work!

5. Add a base to both sides that is the base of the log; if you have an “ln” in the problem, use base “\(e\)”. You’ll typically use this when you have logs in the equation; you can use this instead of using the log loop. Some of these will look familiar!

Solving Logs: Add a Base to Both Sides

 \(\displaystyle {{\log }_{8}}x=-\frac{1}{3}\)

 

\(\require{cancel}  \displaystyle \begin{align}{{\cancel{8}}^{{\cancel{{{{{\log }}_{8}}}}x}}}&={{8}^{{-\frac{1}{3}}}}\\x&=\frac{1}{2}\end{align}\)

 \(\displaystyle \ln \left( {x-4} \right)=5\)

 

\(\displaystyle \begin{align}{{\cancel{e}}^{{\cancel{{\ln }}\left( {x-4} \right)}}}&={{e}^{5}}\\x-4&={{e}^{5}}\\x&={{e}^{5}}+4\approx 152.413\end{align}\)

 \({{\log}_{4}}6+{{\log }_{4}}x-{{\log }_{4}}2=3\)

 

\(\begin{align}{{\log }_{4}}\left( {\frac{{6x}}{2}} \right)&=3\\{{\cancel{4}}^{{\cancel{{{{{\log }}_{4}}}}\left( {\frac{{6x}}{2}} \right)}}}&={{4}^{3}}\\3x&=64\\x&=\frac{{64}}{3}\end{align}\)

 \({{\log }_{5}}\left( {{{x}^{2}}+4x+4} \right)=2\)

 

\(\begin{align}{{\cancel{5}}^{{\cancel{{{{{\log }}_{5}}}}\left( {{{x}^{2}}+4x+4} \right)}}}&={{5}^{2}}\\{{x}^{2}}+4x+4&=25\\{{x}^{2}}+4x-21&=0\\\left( {x+7} \right)\left( {x-3} \right)&=0\\x=-7,\,\,\,\,x=3\end{align}\)

 

Both work!

Solving Log Equations

Let’s do some problems and see what techniques we use:

Solving Using Logs

Notes

 \({{\log }_{2}}\left( {2{{n}^{2}}+10} \right)={{\log }_{2}}\left( {22-5n} \right)\)

 

\(\begin{align}2{{n}^{2}}+10&=22-5n\\2{{n}^{2}}+5n-12&=0\\\left( {2n-3} \right)\left( {n+4} \right)&=0\\n=\frac{3}{2},\,\,\,\,n=-4\end{align}\)

If the same log and same base are on both sides, you can just set arguments of logs equal to each other (one-to-one property).

 

Since we have a quadratic, we have to set to 0 and see if we can factor; if not, use the quadratic formula (we can factor).

 

Remember to check your answers to make sure you don’t have a negative log argument. Both work!     √

 \({{2}^{{n+3}}}=128\)

 

\(\begin{align}\ln \left( {{{2}^{{n+3}}}} \right)&=\ln \left( {128} \right)\\\left( {n+3} \right)\ln \left( 2 \right)&=\ln \left( {128} \right)\,\\n+3&=\frac{{\ln \left( {128} \right)}}{{\ln \left( 2 \right)}}\\n&=4\end{align}\)    or \(\require{cancel} \displaystyle \begin{align}\cancel{{{{{\log }}_{2}}}}\left( {{{{\cancel{2}}}^{{n+3}}}} \right)&={{\log }_{2}}\left( {128} \right)\,\\n+3&={{\log }_{2}}\left( {128} \right)\\\,n+3&=\frac{{{{{\log }}_{2}}\left( {128} \right)}}{{{{{\log }}_{2}}\left( 2 \right)}}\\\,n&=4\end{align}\)

Use the power rule to get the exponent down if the variable is in the exponent.

 

Take the ln of both sides; we can use ln since we we’re not dealing with base 10. (I also show how to use \({{\log }_{2}}\) to get the same answer).

 

Remember to check your answer. It works!     √

 \(2\ln \left( x \right)=3\)

 

\(\begin{align}\ln \left( x \right)&=\frac{3}{2}\\{{e}^{{\frac{3}{2}}}}&=x\\x\,\,&\approx \,\,4.482\end{align}\)

Use the loop to get the variable out of the log argument. You typically use this if you have logs in the equations.

 

We first need to divide both sides by 2 to get the log (ln) by itself.  Then use the loop to get the \(x\) out of the ln argument.

We can use the calculator to get the answer or just leave it in exact form (\({{e}^{{\frac{3}{2}}}}\)). We have a positive log argument, so it works.

 \({{\log }_{2}}\left( {2n+6} \right)-{{\log }_{2}}\left( n \right)=3\)

 

\(\begin{align}{{\log }_{2}}\left( {\frac{{2n+6}}{n}} \right)=3;\\{{\cancel{2}}^{{\cancel{{{{{\log }}_{2}}}}\left( {\frac{{2n+6}}{n}} \right)}}}={{2}^{3}};\,\,\,\frac{{2n+6}}{n}=8\\2n+6=8n;\,\,\,n=1\end{align}\)

Use the quotient rule to condense the logs and then raise both sides to a base that is the base of the log.

 

Instead of using the loop, we can just raise both sides to the base and eliminate the log this way (remember that \({{a}^{{{{{\log }}_{a}}x}}}=x\)).

 

Remember to check your answer to make sure you don’t have a negative log argument. It works!     √

 \(5{{e}^{{.004x}}}=35\)

 

\(\displaystyle \begin{array}{c}{{e}^{{.004x}}}=7\\\cancel{{\ln }}{{\cancel{e}}^{{.004x}}}=\ln \left( 7 \right)\,\,;\,\,\,\,.004x=\ln \left( 7 \right)\end{array}\)

\(\displaystyle x=\frac{{\ln \left( 7 \right)}}{{.004}}\,\,\approx \,\,486.48\)

Use the power rule to get the exponent down if the variable is in the exponent.

 

First divide by 5 to get \({{e}^{{.004x}}}\) by itself. Then take the ln of each side and use the power rule to get the term with the variable down. We can typically keep the answer in exact form (with the ln).

 

Remember to check your answer. It works!     √

Here are a few more:

Solving Using Logs

Notes

 \({{\log }_{{2x}}}6=.5\)

 

\(\begin{align}{{\left( {2x} \right)}^{{.5}}}&=6\\{{\left[ {{{{\left( {2x} \right)}}^{{.5}}}} \right]}^{2}}&={{6}^{2}}\\2x&=36\\x&=18\end{align}\)

Use the loop to get the exponent down. You typically use this if you have logs in the equations.

 

Watch the parentheses; we have to raise \(2x\) to the .5, and then it’s easiest to square both sides.

This time we had the variable in the base, so we just raised both sides to the reciprocal of the exponent to get rid of it.

 

The answer works!  √

 \(\displaystyle {{.7}^{{\frac{{50}}{x}}}}=28\)

 

\(\displaystyle \begin{align}\ln \left( {{{{.7}}^{{\frac{{50}}{x}}}}} \right)&=\ln \left( {28} \right)\\\left( {\frac{{50}}{x}} \right)\ln \left( {.7} \right)&=\ln \left( {28} \right)\\\frac{{50}}{x}&=\frac{{\ln \left( {28} \right)}}{{\ln \left( {.7} \right)}}\,\,\approx \,\,-9.3424\\\cancel{x}\cdot \frac{{50}}{{\cancel{x}}}&=-9.3424\cdot x\\x&=\frac{{50}}{{-9.3424}}\,\,\approx \,\,-5.35\end{align}\)

Use the power rule to get the exponent down if the variable is in the exponent.

 

Take the ln of both sides; we can use ln since we we’re not dealing with base 10.

When we get \(\displaystyle \frac{{50}}{x}\) on one side, we have to either cross multiply or multiply by \(x\) on both sides, and then divide by –9.3424 to get \(x\).

 

Remember to check your answer. It works!     √

\(\displaystyle {{\log }_{4}}{{\left( {2x-5} \right)}^{5}}=20\)

 

\(\displaystyle \begin{align}5{{\log }_{4}}\left( {2x-5} \right)&=20\\{{\log }_{4}}\left( {2x-5} \right)&=4\\{{4}^{4}}&=2x-5\\2x&=256+5\\2x&=261\\x&=\frac{{261}}{2}=130.5\end{align}\)

First, use the power rule to get the power down; it will be easier to solve this way (especially without a calculator!).

 

Use the loop to get the variable out of the log argument. We could have also made each side an exponent with base 4: \(\displaystyle {{4}^{{{{{\log }}_{4}}\left( {2x-5} \right)}}}={{4}^{4}}\), and set the exponents equal to each other.

 

The answer works!    √

Applications of Logs

Again, we typically use logs to solve problems where we have a variable in the exponent; we can use the Power Rule to “get the exponent down”.

There are basically two formulas used for exponential growth and decay, and when we need to solve for any variables in the exponents, we’ll use logs. These formulas are \(A=P{{\left( 1+\frac{r}{n} \right)}^{nt}}\) and \(A=P{{e}^{rt}}\), which is also written in these types of problems as \(A=P{{e}^{kt}}\).

Let’s first redo the problem earlier to see how much easier it is to use logs than “guess and check” when the variable is in the exponent.

Problem:
Suppose that a graduating class had 500 students graduating the first year, but after that, the number of students graduating declines by a certain percentage.

  (a)  If the number of students graduating will be 400 in 2 years, what is the decay rate?
  (b)  Using this same decay rate, in about how many years will there be less than 300 students?

Solution:
(a)  We got the decay rate (10.6%, or .106) by working the here in the Exponential Functions section.

(b)  Now we can finally use logs instead of “Guess and Check”:

Math Using Logs Notes

Suppose that a graduating class had 500 students graduating the first year, but after that, the number of students graduating declines by a certain percentage.

(a)  If the number of students graduating will be 400 in 2 years, what is the decay rate?

(b) Using this same decay rate, in about how many years will there be less than 300 students?

 

\(\begin{align}A&=P{{\left( {1-r} \right)}^{t}}\\300&=500{{\left( {1-.106} \right)}^{t}}\\\frac{{300}}{{500}}&={{\left( {.894} \right)}^{t}}\\\ln \left( {.6} \right)&=\ln {{\left( {.894} \right)}^{t}}\\\ln \left( {.6} \right)&=t\ln \left( {.894} \right)\\t&=\frac{{\ln \left( {.6} \right)}}{{\ln \left( {.894} \right)}}\approx 4.56\end{align}\)

(a) We’ll use the exponential decay formula, where \(P\) (starting number) is 500, the decay rate is .106 from (a), and we need to find \(t\) where \(A\) (ending number) is 300. The decay factor is .894 (1 – .106).

 

(b) Now, instead of “guessing and checking” for the value in the exponent, we can use logs. We get the expression with the exponent by itself, and then take the ln of both sides. We can then use the Power Rule to “get the exponent down”.

 

In about 4.56 years, there will be less than 300 students. That’s the same answer we got with the “Guess and Check” problem above. Using logs is much faster and more accurate

Population Growth Problem

Many times problems will give you the exponential formula, and you basically have to plug in to get the answers:

A population of 50 wolves was introduced into a forest in 2009. The population is expected to grow by the function \(p\left( t \right)=50{{e}^{.085t}},\) where \(t\) is in years.   

  (a)  What will be the population in 2015?  
  (b)  How many years will it take for the population to double?  
  (c)  During what year will the population double?

Solution:
We have to be a little careful when given dates like in this problem. Typically (and unless otherwise stated), \(t\) represents the number of years after the first year, in our case, 2009.

Math Notes

A population of 50 wolves was introduced into a forest in 2009. The population is expected to grow by the function \(p\left( t \right)=50{{e}^{.085t}}\), where \(t\) is in years.

(a) What will be the population in 2015?  

 

\(\displaystyle \begin{align}p\left( t \right)&=50{{e}^{{.085t}}}\\p\left( 6 \right)&=50{{e}^{{.085\left( 6 \right)}}}\\&\approx 83.26\\&=83\text{ wolves}\end{align}\)

(a) To get the population in year 2015, we can get \(t\) by subtracting 2009 from 2015: to get \(t=6\). Plug in 6 for \(t\) to get 83 wolves in the year 2015. (We can’t have .26 of a wolf!)

(b) How many years will it take for the population to double?  

 

\(\displaystyle \begin{align}100&=50{{e}^{{.085t}}}\\2&=1{{e}^{{.085t}}}\\\ln \left( 2 \right)&={{\cancel{{\ln e}}}^{{.085t}}}\\\ln \left( 2 \right)&=.085t\\t&=\frac{{\ln \left( 2 \right)}}{{.085}}\approx 8.15\end{align}\)

(b) We’ll use the same formula, but now we need to use logs to get the variable in the exponent down.

When the wolf population doubles, we’ll have 100 wolves. Plug in 100 for “\(y\)” or \(P(t)\) and work backwards to get \(t\).

 

We get the expression with the exponent by itself, and then take the ln of both sides. (Always use ln when you have a formula with \(e\) in it). In this case, the ln and \(e\) sort of cancel themselves out.

 

In about 8.15 years, the wolf population will double.

(c) During what year will the population double?
(c) Since the wolf population will double in a little over 8 years, during the year  (2009 + 8), the wolf population will double.

Continuous Compounding Problem:

Madison really wants to buy a car. She needs a down payment of $4000. If she deposits $3500 now with interest compounding continuously at 3%, how many years will it take her to save enough to buy the car?

Solution:

Math Using Logs Notes

Madison really wants to buy a car. She needs a down payment of $4000. If she deposits $3500 now with interest compounding continuously at 3%, how many years will it take her to save enough to buy the car?

 

\(\require{cancel} \begin{align}A&=P{{e}^{{rt}}}\\4000&=3500{{e}^{{(.03)t}}}\\\frac{8}{7}&={{e}^{{(.03)t}}}\\\ln \left( {\frac{8}{7}} \right)&={{\cancel{{\ln e}}}^{{(.03)t}}}\\t&=\frac{{\ln \left( {\frac{8}{7}} \right)}}{{.03}}\approx 4.45\end{align}\)

We’ll use the formula for compounding interest, where \(P\) (the beginning value) is $3500, and \(A\) (the ending value) is $4000. The interest rate is .03:

 

We will use the “PERT” equation, since we’re talking about compounding interest. Remember that you “continuously” wash your hair with “Pert” shampoo.

 

We get the expression with the exponent by itself, and then take the ln of both sides. (Always use ln when you have a formula with \(e\) in it). In this case, the ln and \(e\) sort of cancel themselves out.

 

In about 4.45 years, Madison will have enough for the down payment for her car.

Using Logs to Find the Rate Problem:

Aven wants to buy a car in 4 years and needs a down payment of $2500. If she deposits $2000 now, with interest compounded continuously, what interest rate will she need to get her down payment in time?

Solution:

Math Using Logs Notes

Aven wants to buy a car in 4 years and needs a down payment of $2500. If she deposits $2000 now, with interest compounded continuously, what interest rate will she need to get her down payment in time?

 

\(\begin{align}A&=P{{e}^{{rt}}}\\2500&=2000{{e}^{{r(4)}}}\\1.25&={{e}^{{4r}}}\\\ln \left( {1.25} \right)&={{\cancel{{\ln e}}}^{{4r}}}\\t&=\frac{{\ln \left( {1.25} \right)}}{4}\approx .056\\&\approx 5.6\%\end{align}\)

We’ll use the formula for compounding interest again, where \(P\) (the beginning value) is $2000, \(P\) (the ending value) is $2500, and \(t\) (number of years) is 4. We need to use logs to find the interest rate, since it is in the exponent.

 

We get the expression with the exponent by itself, and then take the ln of both sides. (Always use ln when you have a formula with \(e\) in it). In this case, the ln and \(e\) sort of cancel themselves out.

 

The interest rate will have to be about 5.6% for Aven to have enough money in 4 years for her down payment.

Some problems require a two-part solution, where first we solve for the exponential growth or decay rate – typically the \(k\), and then we solve again, with the \(k\) in the problem.

Problem:

A large colony of fleas is growing exponentially on the family dog (yuck!). There are 400 fleas initially.  
  (a)  If there are 600 fleas after 1 day, how many will there be after 5 days?  
  (b)  How long will it be until there are 10,000 fleas?

Solution:

Log Problem Using Calculator Notes

A large colony of fleas is growing exponentially on the family dog (yuck!). There are 400 fleas initially.

 

\(\require{cancel} \displaystyle \begin{align}A&=P{{e}^{{kt}}}\\600&=400{{e}^{{k\left( 1 \right)}}}\\1.5&={{e}^{k}}\\\ln \left( {1.5} \right)&={{\cancel{{\ln e}}}^{k}}\\k&=\ln \left( {1.5} \right)\approx .4054651081\end{align}\)

We need to find \(k\) first by using the initial number of fleas (400), and the number of fleas after 1 day (600). Get the \({{e}^{k}}\) by itself, and then take the ln of both sides.

 

We need to use a \(k\) that is .4054651081. We have to use the most accurate \(k\) we can, since logs are sensitive. It’s best to keep it in your calculator if you can – you can use the STORE function in your calculator:

              Now that we have \(k\), we can use it to answer (a) and (b):
 

(a) If there are 600 fleas after 1 day, how many will there be after 5 days?

 

\(\begin{align}A&=P{{e}^{{kt}}}\\A&=400{{e}^{{\left( {.4054651081} \right)\left( 5 \right)}}}\\A&\approx 3037\end{align}\)

(a) We still initially have 400 fleas on the dog. But now we know \(k\), and we use it to find the number of fleas after 5 days. If \(k\) is stored in \(x\), you can use your calculator:

There will be about 3037 fleas after 5 days.

(b) How long will it be until there are 10,000 fleas?

 

\(\displaystyle \begin{align}A&=P{{e}^{{\left( {.4054651081} \right)t}}}\\10000&=400{{e}^{{\left( {.4054651081} \right)t}}}\\25&={{e}^{{\left( {.4054651081} \right)t}}}\\\ln \left( {25} \right)&={{\cancel{{\ln e}}}^{{\left( {.4054651081} \right)t}}}\\t&=\frac{{\ln \left( {25} \right)}}{{.4054651081}}\approx 7.9387\end{align}\)

(b) Now, use the formula again with the \(k\) that we found earlier to find how many days it will take for the dog to have 10000 fleas. Again, if \(k\) is stored in \(x\), you can use your calculator:

 

In about 8 days there will be 10,000 fleas. Yikes!

 

It’s always a good idea to check your work by putting the \(t\) that you get back in the original equation – it works! Again, you can use the store function in your calculator to do this.  √

Revisiting Half Life Problem

We can solve half-life problems using two different methods; we’ll use both methods here. We solved a half-life problem above in the Exponents section, but if you need to find a time (a variable in the exponent), then you need to use logs.

Remember that half-life problems deal with exponential decays that halve for every time period. For example, if we start out with 20 grams, after the next time period, we’d have 10, then 5, and so on. For these problems, the base (decay factor) of the exponential equation is .5.

The trick on half-life problems is to raise the .5 to \(\displaystyle \frac{{\text{time period we want}}}{{\text{time for one half-life}}}\), since this will give us the number of times the substance actually halves.

Solution (Method 1, like we used in the Exponential Functions section):

Half-Life Problem Math Notes
A chemical substance has a half-life of 6 hours.

 

 

(a) How much of a 40-gram sample remains after 18 hours?

\(\begin{array}{c}y=a{{b}^{{\frac{t}{p}}}}\\\\y=40{{\left( {.5} \right)}^{{\frac{{18}}{6}}}}\\y=5\,\,\text{grams}\end{array}\) (a) For this problem, we need to use the exponential equation \(\displaystyle y=a{{b}^{{\frac{t}{p}}}}\), since we have a time period for which the substance divides in two.

 

Since we’re dealing with a half-life problem, we know the decay factor is .5, since it halves every 6 hours. Since the half-life (decay interval) is 6, but we want to know how much of the substance there will be after 18 hours, our “time” is actually\(\displaystyle \frac{{18}}{6}\,\) or \(\displaystyle \frac{{\text{time we are interested in}}}{{\text{time of a half-life}}}\), which is 3.

 

Since we start with 40 grams, \(a=40\). Using the formula, we get \(y=5\) grams.

 

(b) How long until there are only 2 grams left?

 

\(\displaystyle \begin{align}y&=a{{\left( {.5} \right)}^{{\frac{t}{p}}}}\\2&=\left( {40} \right){{\left( {.5} \right)}^{{\frac{t}{6}}}}\\.05&={{.5}^{{\frac{t}{6}}}}\\\ln \left( {.05} \right)&=\ln {{\left( {.5} \right)}^{{\frac{t}{6}}}}\\\ln \left( {.05} \right)&=\frac{t}{6}\ln \left( {.5} \right)\\t&=\frac{{\ln \left( {.05} \right)}}{{\ln \left( {.5} \right)}}\left( 6 \right)\\&\approx 25.93\end{align}\) (b) We’ll need to use logs to find the time until there are only 2 grams left. The beginning amount \(a\) is still 40, the ending amount \(y\) is 2, and the half-life \(p\) is 6:

 

We get the expression with the exponent by itself, and then take the ln of both sides. We can then use the Power Rule to “get the exponent down”.

 

So in about 26 hours there will be 2 grams left.

 

It’s always a good idea to make sure your answer makes sense (the amount of 40 grams should divide by 2 about \(\displaystyle \frac{{26}}{6}\) or around 4 times to get down to 2). Also, check your work by putting the \(t\) that you get back in the original equation – it works!  √

Solution (Method 2):

Sometimes you will learn to solve a half-life problem using the \(A=P{{e}^{kt}}\)  formula and get the \(k\) first, like we did in the flea problem earlier. This method seems a little bit more difficult, but sometimes you are asked for the half-life equation with the \(k\) in it.

Also note that sometimes you are given the equation \(A=P{{e}^{-kt}}\)  for half-life or any decay problems, and then the \(k\) will end up being positive instead of negative,since we would already have a negative sign in the exponent.

Half-Life Problem Math Notes
A chemical substance has a half-life of 6 hours.

 

 

(a) How much of a 40-gram sample remains after 18 hours?

\(\require{cancel} \displaystyle \begin{align}A&=P{{e}^{{kt}}}\\1&=2{{e}^{{k(6)}}}\\.5&={{e}^{{k(6)}}}\\\ln \left( {.5} \right)&=\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{k(6)}}}} \right)\\6k&=\ln \left( {.5} \right)\\k&=\frac{{\ln \left( {.5} \right)}}{6}\\&\approx -.1155245301\\\\A&=40{{e}^{{-.1155245301t}}}\\&=40{{e}^{{-.1155245301\left( {18} \right)}}}\\&=5\text{ grams}\end{align}\) (a) We want to solve for \(k\) first in the exponential equation \(A=P{{e}^{{kt}}}\) when the beginning amount is one half of the ending amount, and \(t\) is the half-life.

It’s a good idea to store \(k\) in our calculator again.

(Note that sometimes we see the equation (if it’s a decay) with a negative sign in front of the \(k\), so we get a positive value for \(k\). We will get the same answer.)

 

The half life equation for that situation is \(\displaystyle A=40{{e}^{{-.1155245301t}}}\).

 

To get the amount after 18 hours, we plug in 18 for \(t\). The beginning amount \(A\) is 40. We see again that after 18 hours, we have 5 grams remaining; this is what we got using the other method!

(b) How long until there are only 2 grams left?

 

\(\displaystyle \begin{align}A&=P{{e}^{{kt}}}\\2&=40{{e}^{{-.1155245301t}}}\\.05&={{e}^{{-.1155245301t}}}\\\ln \left( {.05} \right)&=\cancel{{\ln }}\left( {{{{\cancel{e}}}^{{-.1155245301t}}}} \right)\\t&=\frac{{\ln \left( {.05} \right)}}{{-.1155245301}}\\&\approx 25.93\end{align}\) (b) Now we need to use logs again to find the time until there are only 2 grams left.

We will need to use logs to get \(t\). The beginning amount \(P\) is still 40, the ending amount \(A\) is 2, and \(k\) is the same.

 

To get the expression with the exponent by itself, take the ln of both sides and then use the Power Rule to “get the exponent down”.

 

Note that we get the same answer as above: in about 26 hours there will be 2 grams left.

Logarithmic Inequalities

You may have to solve inequality problems (either graphically or algebraically) with logarithmic functions. Remember that we learned about using the Sign Chart or Sign Pattern method for inequalities here in the Quadratic Inequalities section, and also we have the domain restriction that the argument of a log has to be \(>0\).

Let’s do a few inequality problems with exponents and logs:

Log Inequality Notes/Graph
\(\displaystyle 4{{\log }_{2}}\left( {7-x} \right)<8\,\,\,\,\,\,\,\,7-x>0\)

 

\(\displaystyle \begin{align}{{\log }_{2}}\left( {7-x} \right)&<2\,\,\,\,\,\,\,\,\,\,\,x<7\\{{2}^{{{{{\log }}_{2}}\left( {7-x} \right)}}}&<{{2}^{2}}\\7-x&<4\\x&>3\,\,\,\,\text{and}\,\,x<7\\&\left( {3,\,\,7} \right)\end{align}\)

Note that we have to solve the log inequality, but also set the log argument to \(>0\), since all log arguments have to be \(>0\). (This is a domain restriction.)

 

To make sure we keep the inequality signs in the correct place, I found it easiest to raise both sides to the log base, which is 2. Then we have to use the intersection of both inequalities, which is \((3,7)\).

 

You can check your answer by putting both sides of the inequality in your graphing calculator and seeing where the log function \(4{{\log }_{2}}\left( {7-x} \right)\) is underneath the line \(y=8\). It may be hard to see, but the log graph stops at \(x=7\), so that’s the upper bound.

Use the graphing calculator to solve:

 

\({{\log }_{6}}\left( {7x+1} \right)\le {{\log }_{4}}\left( {4x-4} \right)\)

 

 

Using the graphing calculator, put use the logs in \({{Y}_{1}}\) and \({{Y}_{2}}\) using LOGBASE (MATH A or ALPHA, WINDOW 5).

After graphing the two expressions, hit 2nd TRACE (CALC) 5 to get the intersection (before that, make sure the cursor is near the intersection by using TRACE and arrows). After “First Curve?”, move cursor up to first log curve using the arrow keys (if it’s not there) and hit ENTER. Do the same for the other log curve after “Second Curve?”. Then hit ENTER after “Guess?” to get the intersection.

 

We can see that the blue graph is below the red graph when \(x\ge 5\), so the answer is \(\left[ {5,\,\,\infty } \right)\).

Learn these rules, and practice, practice, practice!


For Practice: Use the Mathway widget below to try a Condensing a Log problem. Click on Submit (the blue arrow to the right of the problem) and click on Write as a Single Logarithm to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Solving Inequalities – you are ready!

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