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Section 15.1 Evaluating Logarithmic Functions

Example 15.1. Exploring and Motivation.

Suppose \(f(x)=x+3\text{.}\) What is \(f^{-1}(x)\text{?}\) Well, we know from Section 6.2 how to do this:

\begin{align*} y \amp= x+3\\ x \amp= y+3\\ x-3 \amp= y \end{align*}

So, we have \(f^{-1}(x)=x-3\text{.}\) This makes sense, since subtraction is the opposite of addition, it should be the inverse.

Let's do another. What is the inverse of \(g(x)=3x\text{?}\) Again, we know how to do this:

\begin{align*} y \amp= 3x\\ x \amp= 3y\\ \frac{x}{3} \amp= y \end{align*}

So, we have \(g^{-1}(x)=\frac{x}{3}\text{.}\) This makes sense, since division is the opposite of multiplication, it should be the inverse.

One more: what is the inverse of \(h(x)=x^3\text{?}\) You know the drill by now:

\begin{align*} y \amp= x^3\\ x \amp= y^3\\ \sqrt[3]{x} \amp= y \end{align*}

So, we have \(h^{-1}(x)=\sqrt[3]{x}\text{.}\) This makes sense, since the cube root is the opposite of a cube, it should be the inverse.

So what do we do with something like \(3^x\text{?}\) This is an exponential function, so we need something that is the opposite of an exponential. This is where logarithms come in.

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Definition 15.2.

If \(f(x)=b^x\) is an exponential function with base \(b\text{,}\) then its invserse is \(f^{-1}(x)=\log_b(x)\text{,}\) the logarithm base \(b\).

Note: inverses work in both directions. That means that if \(g(x)=\log_b(x)\text{,}\) then \(g^{-1}(x)=b^x\text{.}\)

Example 15.3.

Suppose \(f(x)=5^x\) and we want to find its inverse. Since this is an exponential function, its inverse will be a logarithm. The bases need to match, so \(f^{-1}(x)=\log_5(x)\text{.}\)

Suppose \(f(x)=\log_7(x)\) and we want to find its inverse. Since this is an logarithm function, its inverse will be an exponential. The bases need to match, so \(f^{-1}(x)=7^x\text{.}\)

Checkpoint 15.4.

Find the inverse of each function below.

  1. \(\displaystyle f(x)=9^x\)

  2. \(\displaystyle f(x)=1.08^x\)

  3. \(\displaystyle f(x)=\log_2(x)\)

  4. \(\displaystyle f(x)=\log_{0.87}(x)\)

Answer.

  1. \(\displaystyle f^{-1}(x)=\log_9(x)\)

  2. \(\displaystyle f^{-1}(x)=\log_{1.08}(x)\)

  3. \(\displaystyle f^{-1}(x)=2^x\)

  4. \(\displaystyle f^{-1}(x)=0.87^x\)

Solution.

  1. Since \(f(x)=9^x\) is an exponential function, its inverse will be a logarithm. The bases need to match, so \(f^{-1}(x)=\log_9(x)\text{.}\)

  2. Since \(f(x)=1.08^x\) is an exponential function, its inverse will be a logarithm. The bases need to match, so \(f^{-1}(x)=\log_{1.08}(x)\text{.}\)

  3. Since \(f(x)=\log_2(x)\) is a logarithm, its inverse will be an exponential function. The bases need to match, so \(f^{-1}(x)=2^x\text{.}\)

  4. Since \(f(x)=\log_{0.87}(x)\) is a logarithm, its inverse will be an exponential function. The bases need to match, so \(f^{-1}(x)=0.87^x\text{.}\)

So know that we know where logarithms come from, how do we actually evaluate them? The key is that inverses cancel. So, the logarithm will cancel with an exponential, as long as the bases match.

Example 15.5.

Let's evaluate each of the following expressions.

  1. Suppose we want to evaluate \(\log_7\left(7^{13}\right)\text{.}\) Since the log is base 7, and so is the exponential, so they cancel and we get that

    \begin{equation*} \log_7\left(7^{13}\right) = 13 \end{equation*}

  2. Suppose we want to evaluate \(\log_4\left(4^{1.072}\right)\text{.}\) Since the log is base 4, and so is the exponential, so they cancel and we get that

    \begin{equation*} \log_4\left(4^{1.072}\right)=1.072 \end{equation*}

  3. Suppose we want to evaluate \(6^{\log_6(17)}\text{.}\) Since the log is base 6, and so is the exponential, so they cancel and we get that

    \begin{equation*} 6^{\log_6(17)}=17 \end{equation*}

  4. Suppose we want to evaluate \(\log_3\left(9\right)\text{.}\) At first glance, the inside is not an exponential with the right base (3). But, all hope is not lost, since 9 is actually a power of 3. Since \(9=3^2\text{,}\)

    \begin{equation*} \log_3(9)=\log_3\left(3^2\right) = 2 \end{equation*}

Sometimes, logarithms can't be simplified any more than they already are. Just like \(\sqrt{25}=5\text{,}\) but \(\sqrt{24}\) is just \(\sqrt{24}\) since \(24\) isn't a perfect square, we have that \(\log_4(16)=2\text{,}\) but \(\log_4(15)\) is just \(\log_4(15)\text{,}\) since \(15\) isn't a power of 4.

Checkpoint 15.6.

Simplify each of the following, if possible

  1. \(\displaystyle \log_8\left(8^3\right)\)

  2. \(\displaystyle 3^{\log_3(5)}\)

  3. \(\displaystyle \log_2\left(8\right)\)

  4. \(\displaystyle \log_7\left(9\right)\)

Solution.

Simplify each of the following, if possible

  1. \(\log_8\left(8^3\right)\text{:}\) Since the log is base 8, and so is the exponential, so they cancel and we get that

    \begin{equation*} \log_8\left(8^3\right) = 3 \end{equation*}

  2. \(3^{\log_3(5)}\text{:}\) Since the log is base 3, and so is the exponential, so they cancel and we get that

    \begin{equation*} 3^{\log_3(5)}=5 \end{equation*}

  3. \(\log_2\left(8\right)\text{:}\) At first glance, the inside is not an exponential with the right base (2). But, all hope is not lost, since 8 is actually a power of 2. Since \(8-2^3\text{,}\)

    \begin{equation*} \log_2(8)=\log_2\left(2^3\right) = 3 \end{equation*}

  4. \(\log_7\left(9\right)\text{.}\) Since 9 is not a power of 7, this one cannot be simplified any further.