This section covers:
 Transformations of Exponential and Log Functions
 Writing Exponential and Logarithmic Functions from a Graph
 Inverses and Compositions of Exponential and Logarithmic Functions
 Exponential and Logarithmic Inequalities
Let’s tie up a few loose ends with Exponential and Logarithmic functions. We’ve covered Transformations of Functions, Inverses of Functions, Compositions of Functions, Radical Inequalities and Quadratic Inequalities in other sections, but let’s specifically focus on exponents and logs.
Transformations of Exponential and Log Functions
The critical or significant points of the parent exponential function \(y={{b}^{x}}\) are \(\left( {1,\,\frac{1}{b}} \right),\,\,\left( {0,1} \right),\,\,\left( {1,b} \right)\), and for the parent logarithmic function \(y={{\log }_{b}}x\) (inverse of exponential function) are \(\left( {\frac{1}{b},\,1,\,} \right),\,\,\left( {1,0} \right),\,\,\left( {b,1} \right)\).
I always remember that the key point of an exponential function is \(\left( {0,\,1} \right)\), since the “e” in exp is round like a “0”, and the key point of a log function is \(\left( {1,\,0} \right)\), since this looks like the LO in “log”. We have to also remember that if the function shifts, this “anchor point” will move.
Remember that 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.
Here are some examples that we saw in the Parent Graphs and Transformations Section:
Writing Exponential and Logarithmic Equations from a Graph
Writing Exponential Equations from Points and Graphs
You may be asked to write exponential equations, such as the following:
 Write an equation to describe the exponential function in form \(y=a{{b}^{x}}\), with a given base and a given point.
 Write an exponential function in form \(y=a{{b}^{x}}\) whose graph passes through two given points. (You may be able to do this using Exponential Regression.)
 For a certain graph, write the appropriate exponential function of the form \(y=a{{b}^{x}}+d\) (an exponential function with a vertical shift).
Let’s try these types of problems:
Problem  Solution 
Write an equation to describe the exponential function in form \(y=a{{b}^{x}}\), with base 3 and passing through the point \(\left( {4,\,\,162} \right)\).  The equation will be in the form \(y=a{{\left( 3 \right)}^{x}}\), since the base is 3. Plug in 4 for x and 162 for y, and solve for a: \(\begin{align}y&=a{{\left( 3 \right)}^{x}}\\162&=a{{\left( 3 \right)}^{4}}\\a&=\frac{{162}}{{81}}=2\end{align}\) So the equation is \(y=2{{\left( 3 \right)}^{x}}\). 
Write an exponential function in form \(y=a{{b}^{x}}\) whose graph passes through the two points \(\left( {3,\,\,10} \right)\) and \(\left( {5,\,\,40} \right)\).  By plugging in the given points, the two equations we’ll have are \(10=a{{\left( b \right)}^{3}}\) and \(40=a{{\left( b \right)}^{5}}\). We need to find a and b; this is a system of equations. The trick is to solve the first equation for \({{b}^{3}}\), and then substitute this in the second equation by factoring the \({{b}^{5}}\) to make \({{b}^{3}}\cdot {{b}^{2}}\). We could have also just solved for b, but this way is easier: \(\require{cancel} \begin{array}{l}10=a{{b}^{3}};\,\,\,\,\,{{b}^{3}}=\frac{{10}}{a}\\40=a{{b}^{5}};\,\,\,\,40=a{{b}^{3}}{{b}^{2}}\\40=\cancel{a}\left( {\frac{{10}}{{\cancel{a}}}} \right){{b}^{2}};\,\,\,\,\,{{b}^{2}}=\frac{{40}}{{10}}=4\\b=2\,\,\,\,\text{(base can }\!\!’\!\!\text{ t be negative)}\end{array}\) \(\begin{array}{c}\text{Plug }b\text{ in either equation for }a:\\10=a{{\left( 2 \right)}^{3}}\\a=\frac{{10}}{8}=\frac{5}{4}\end{array}\)
So the exponential function is: \(y=\frac{5}{4}{{\left( 2 \right)}^{x}}\). Check your points; they work! √ 
For the following graph, write the appropriate exponential function in the form \(y=a{{b}^{x}}+d\) (vertically shifted exponential function):  We see that this graph has an asymptote at \(y=3\), so it will have a vertical shift of –3, or \(d=3\). Our equation will be in the form \(y=a{{b}^{x}}3\).
Now we need to use a system of equations with the two points on the graph: \(\left( {0,1} \right)\) and \(\left( {1,\,1} \right)\). When you have a problem like this, first use any point that has a “0” in it if you can; it will be easiest to solve the system.
Solve for a first using \(\left( {0,1} \right)\): \(\begin{array}{c}1=a{{b}^{0}}3;\,\,\,\,\,a\left( 1 \right)=1+3;\,\,\,\,a=4\\y=4{{b}^{x}}3\end{array}\)
Use this equation and plug in \(\left( {1,\,1} \right)\) to solve for b: \(1=4{{b}^{1}}3;\,\,\,\,\,4b=2;\,\,\,\,\,\,b=.5\)
So the exponential function is \(y=4{{\left( {.5} \right)}^{x}}3\). Graph this function and it works! √

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 an exponential function in form \(y=a\log \left( {xh} \right)+k\) from a graph (given asymptote and two points).
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 \(y=a\log \left( {xh} \right)+k\):
 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 \(y=a\log \left( {x3} \right)+k\).
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 \(\left( {4,2} \right)\):
\(\displaystyle \begin{array}{l}y=a\log \left( {x3} \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=a\log \left( {x3} \right)+2\\2=a\log \left( {43} \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=a\log \left( {133} \right)+2\\2=a\log \left( 1 \right)+k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=a\log \left( {10} \right)+2\,\,\,\,\\2=a\cdot 0+k;\,\,\,\,\,k=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=a\cdot 1+2;\,\,\,\,\,a=2\end{array}\)
So the logarithmic function is \(y=2\log \left( {x3} \right)+2\). Graph this function and it works! √

Note: we may also be able to use logarithmic regression to find logs equations based on points, like we did here with Exponential Regression.
Inverses and Compositions of Exponential and Logarithmic Functions
As it turns out, exponential functions are inverses of log functions and of course vice versa! Let’s show algebraically that the parent exponential and log functions (\(y={{b}^{x}}\,\,and\,\,y={{\log }_{b}}\)) are inverses – three different ways. Again, we learned about how to find Inverses in the Inverses of Functions section.
Here are the graphs of the two functions again, so you can see that they are inverses; note symmetry around the line y = x. Also note that their domains and ranges are reversed:
Let’s find the inverses of the following transformed exponential and log functions by switching the x and the y and solving for the “new” y:
Exponential and Logarithmic Inequalities
You may have to solve inequality problems (either graphically or algebraically) with either exponential or 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:
Learn these rules, and practice, practice, practice!
On to Solving Inequalities – you are ready!