## Section2.5Logarithmic Functions

Recall the three kinds of exponential functions $f(x)=a^x$ depending on whether $0\lt a\lt 1\text{,}$ $a=1$ or $a>1\text{:}$

So long as $a\neq 1\text{,}$ the function $f(x)=a^x$ satisfies the Horizontal Line Test and therefore has an inverse. We call the inverse of $a^x$ the logarithmic function with base a and denote it by $\log_a\text{.}$ In particular,

\begin{equation*} \log_a x=y\iff a^y=x\text{.} \end{equation*}

The cancellation formulas for logs are:

\begin{equation*} \log_a(a^x)=x,\mbox{ for every $x\in\R$}\text{,} \end{equation*}
\begin{equation*} a^{\log_a(x)}=x,\mbox{ for every $x>0$}\text{.} \end{equation*}

Since the function $f(x)=a^x$ for $a\neq 1$ has domain $\mathbb{R}$ and range $(0,\infty)\text{,}$ the logarithmic function has domain $(0,\infty)$ and range $\mathbb{R}\text{.}$ For the most part, we only focus on logarithms with a base larger than $1$ (i.e., $a>1$) as these are the most important.

Notice that every logarithm passes through the point $(1,0)$ in the same way that every exponential function passes through the point $(0,1)\text{.}$

Some properties of logarithms are as follows.

###### Logarithm Properties.

Let $A,B$ be positive numbers and $b>0$ ($b\neq1$) be a base.

• $\ds{\log_b(AB)=\log_b A+\log_b B}\text{,}$

• $\ds{\log_b\left(\frac{A}{B}\right)=\log_b A-\log_b B}\text{,}$

• $\ds{\log_b(A^n)=n\log_b A}\text{,}$ where $n$ is any real number.

###### Example2.21. Compute Lorarithms.

To compute $\log_2(24)-\log_2(3)$ we can do the following:

\begin{equation*} \log_2(24)-\log_2(3)=\log_2\left(\frac{24}{3}\right)=\log_2(8)=3\text{,} \end{equation*}

since $2^3=8\text{.}$

### Subsection2.5.1The Natural Logarithm

As mentioned earlier for exponential functions, the number $e\approx 2.71828\ldots$ is the most convenient base to use in Calculus. For this reason we give the logarithm with base $e$ a special name: the natural logarithm. We also give it special notation:

\begin{equation*} \log_ex=\ln x\text{.} \end{equation*}

You may pronounce $\ln$ as either: “el - en”, “lawn”, or refer to it as “natural log”. The above properties of logarithms also apply to the natural logarithm.

Often we need to turn a logarithm (in a different base) into a natural logarithm. This gives rise to the change of base formula.

###### Change of Base Formula.
\begin{equation*} \log_ax=\frac{\ln x}{\ln a} \end{equation*}
###### Example2.22. Combine Logarithms.

Write $\ln A+2\ln B -\ln C$ as a single logarithm.

Solution

Using properties of logarithms, we have,

\begin{align*} \ln A+2\ln B -\ln C \amp = \ln A + \ln B^2 - \ln C\\ \amp = \ln (AB^2) - \ln C\\ \amp = \ds{\ln\frac{AB^2}{C}} \end{align*}
###### Example2.23. Solve Exponential Equations using Logarithms.

If $e^{x+2}=6e^{2x}\text{,}$ then solve for $x\text{.}$

Solution

Taking the natural logarithm of both sides and noting the cancellation formulas (along with $\ln e=1$), we have:

\begin{equation*} \begin{array}{rcl} e^{x+2} \amp = \amp 6e^{2x}\\ \\ \ln e^{x+2}\amp =\amp \ln (6e^{2x})\\ \\ x+2\amp =\amp \ln 6 + \ln e^{2x}\\ \\ x+2\amp =\amp \ln 6 + 2x\\ \\ x\amp =\amp 2-\ln 6 \end{array} \end{equation*}
###### Example2.24. Solve Logarithm Equations using Exponentials.

If $\ln(2x-1)=2\ln(x)\text{,}$ then solve for $x\text{.}$

Solution

“Taking $e$” of both sides and noting the cancellation formulas, we have:

\begin{equation*} \begin{array}{rcl} e^{\ln(2x-1)} \amp = \amp e^{2\ln(x)}\\ \\ (2x-1) \amp = \amp e^{\ln(x^2)}\\ \\ 2x-1 \amp = \amp x^2\\ \\ x^2-2x+1 \amp = \amp 0\\ \\ (x-1)^2 \amp = \amp 0 \end{array} \end{equation*}

Therefore, the solution is $x=1\text{.}$

##### Exercises for Section 2.5.

Expand $\ds\log_{10} ((x+45)^7 (x-2))\text{.}$

Expand $\ds\log_2 {\frac{x^3}{3x-5 +(7/x)}}\text{.}$

Write $\ds \log_2 3x + 17 \log_2 (x-2) - 2\log_2 (x^2 + 4x + 1)$ as a single logarithm.

$\log_{2}\left(\dfrac{3x(x-2)^{17}}{(x^{2}+4x+1)^{2}} \right)$
Solution

We can write the expression as a single logarithm using log-rules.

\begin{equation*} \begin{split} \log_{2}3x + 17\log_{2}(x-2) - 2\log_{2}(x^{2}+4x+1) \amp = \log_{2}3x + \log_{2}(x-2)^{17} - \log_{2}(x^{2}+4x+1)^{2} \\ \amp = \log_{2}\left(\dfrac{3x(x-2)^{17}}{(x^{2}+4x+1)^{2}} \right) \end{split} \end{equation*}

Solve $\ds \log_2 (1+ \sqrt{x} ) = 6$ for $x\text{.}$

$x = 3969$
Solution

We solve for $x$ as follows.

\begin{equation*} \begin{split} \amp \log_{2} \left(1+\sqrt{x}\right) = 6\\ \amp 1 + \sqrt{x} = 2^{6} \\ \amp \sqrt{x} = 2^{6}-1 \\ \amp x = \left(2^{6}-1\right)^{2} \end{split} \end{equation*}

Therefore, $x = 3969\text{.}$

Solve $\ds 2^{x^2} = 8$ for $x\text{.}$

Solve $\ds \log_2 (\log_3 (x) ) = 1$ for $x\text{.}$

Solve $\ds a^{2x} - 4a^{x} + 4 = 0$ for $x$ ($a$ constant).

$\ds x = \dfrac{\log{2}}{\log{a}}$