# Exponentials and Logarithms II

Alignments to Content Standards: F-BF.B.4.b

Let $f$ be the function defined by $f(x)=10^x$ and $g$ be the function defined by $g(x)=\log_{10}(x)$.

1. Sketch the graph of $y=f(g(x))$. Explain your reasoning.
2. Sketch the graph of $y=g(f(x))$. Explain your reasoning.
3. Let $f$ and $g$ be any two inverse functions. For which values of $x$ does $f(g(x))=x$? For which values of $x$ does $g(f(x))=x$?

## IM Commentary

The purpose of this task is to help students see that the definition of the logarithm as an exponent results in an inverse relationship between the exponential and logarithm functions with the same base. This prepares them to think about the shape of the graph of the logarithm function in terms of the shape of the graph of the exponential function. As preparation for the reasoning required in this task, students might work through Exponentials and Logarithms I.

## Solution

1. In this case, we are trying sketch $$f(g(x))=10^{\log x}$$ where $log$ denotes the logarithm with base 10. We know that $10^{\log x} = x$ because $\log x$ is, by definition, the exponent which $10$ must be raised to in order to give a result of $x$. Thus we must sketch a graph of $f(g(x)) = x$. The domain of $g(x) = \log x$ is all positive real numbers and this is also the domain of $f(g(x))=10^{\log x}$. Thus our graph should be a graph of $y = x$ with domain $x > 0$: 2. Here we are trying to sketch $$g(f(x)) = \log{10^x}$$ where the logarithm is with base $10$. By definition, $\log{10^x}$ is the exponent $10$ must be raised to in order to give a result of $10^x$ so $$\log{10^x} = x.$$ There is no restraint on the domain of $g(f(x))= \log{10^x}$ because the domain of $f(x)=10^x$ is any real number and its range is all positive real numbers. So all the values of $f(x)$ are in the domain of $g(x)=log_b(x)$. Thus our graph is a graph of $y = x$ with no restriction on the domain: 3. Generalizing what we saw in parts (a) and (b), if $f$ and $g$ are inverse functions then

$g(f(x)) =x$ for all values of $x$ in the domain of $f$ and

$f(g(x)) =x$ for all values of $x$ in the domain of $g$.

For example, in part (a), we see that $f(g(-1))\neq -1,$ because $-1$ is not in the domain of $g$.