Exponential and logarithmetic functions

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Exponential and logarithmic functions illuminated.

Exponential and logarithmic functions may seem somewhat esoteric at first, but they model many phenomena in the real-world.

What are exponential and logarithmic functions?

An exponential function is a function of the form $f(x) = b^x$ where $b\ne 1$ is a positive real number. The domain of an exponential function is $(-\infty ,\infty )$ and the range is $(0, \infty )$ .

Is $b^{-x}$ an exponential function?

yes no

Note that $b^{-x} = \left (b^{-1}\right )^x = \left (\frac {1}{b}\right )^x.$

A logarithmic function is a function defined as follows $\log _b(x) = y \qquad \text {means that}\qquad b^y = x$ where $b\ne 1$ is a positive real number. The domain of a logarithmic function is $(0,\infty )$ and the range is $(- \infty , \infty )$ .

In either definition above $b$ is called the base.

Remember that with exponential and logarithmic functions, there is one very special base: $e = 2.7182818284590\ldots$ This is an irrational number that you will see frequently. The exponential with base $e$ , $f(x) = e^x$ is often called the ‘natural exponential’ function. For the logarithm with base $e$ , we have a special notation, $\ln x$ is ‘natural logarithm’ function. We’ll talk about where $e$ comes from when we talk about derivatives.

Connections between exponential functions and logarithms

Let $b$ be a positive real number with $b\ne 1$ .

$b^{\log _b(x)} = x$ for all positive $x$
$\log _b(b^x) = x$ for all real $x$

What exponent makes the following expression true? $3^x = e^{\left ( x \cdot \answer {\ln 3} \right )}.$

What can the graphs look like?

Graphs of exponential functions

Here we see the the graphs of four exponential functions.

Match the curves $A$ , $B$ , $C$ , and $D$ with the functions $e^x, \qquad \left (\frac {1}{2}\right )^{x}, \qquad \left (\frac {1}{3}\right )^{x}, \qquad 2^{x}.$

One way to solve these problems is to compare these functions along the vertical line $x=1$ ,

Note $\left (\frac {1}{3}\right )^1 < \left (\frac {1}{2}\right )^1 < 2^1 < e^1.$ Hence we see:

$\left (\frac {1}{3}\right )^{x}$ corresponds to $\answer [given]{B}$ .
$\left (\frac {1}{2}\right )^{x}$ corresponds to $\answer [given]{A}$ .
$2^x$ corresponds to $\answer [given]{D}$ .
$e^x$ corresponds to $\answer [given]{C}$ .

Graphs of logarithmic functions

Here we see the the graphs of four logarithmic functions.

Match the curves $A$ , $B$ , $C$ , and $D$ with the functions $\ln (x),\qquad \log _{1/2}(x), \qquad \log _{1/3}(x),\qquad \log _2(x).$

First remember what $\log _b(x)=y$ means: $\log _b(x) = y \qquad \text {means that}\qquad b^y = x.$ Moreover, $\ln (x) = \log _e(x)$ where $e= 2.71828\dots$ . So now examine each of these functions along the horizontal line $y=1$

Note again (this is from the definition of a logarithm) $\left (\frac {1}{3}\right )^1 < \left (\frac {1}{2}\right )^1 < 2^1 < e^1.$ Hence we see:

$\log _{1/3}(x)$ corresponds to $\answer [given]{B}$ .
$\log _{1/2}(x)$ corresponds to $\answer [given]{A}$ .
$\log _2(x)$ corresponds to $\answer [given]{D}$ .
$\ln (x)$ corresponds to $\answer [given]{C}$ .

Properties of exponential functions and logarithms

Working with exponential and logarithmic functions is often simplified by applying properties of these functions. These properties will make appearances throughout our work.

Properties of exponents

Let $b$ be a positive real number with $b\ne 1$ .

$b^m\cdot b^n = b^{m+n}$
$b^{-1} = \frac {1}{b}$
$\left (b^m\right )^n = b^{mn}$

What exponent makes the following true? $2^4 \cdot 2^3 = 2^{\answer {7}}$

$(2^4) \cdot (2^3) = (2 \cdot 2\cdot 2 \cdot 2) \cdot (2 \cdot 2\cdot 2)$

Properties of logarithms

Let $b$ be a positive real number with $b\ne 1$ .

$\log _b(m\cdot n) = \log _b(m) + \log _b(n)$
$\log _b(m^n) = n\cdot \log _b(m)$
$\log _b\left (\frac {1}{m}\right ) = \log _b(m^{-1}) = -\log _b(m)$
$\log _a(m) = \frac {\log _b(m)}{\log _b(a)}$

What value makes the following expression true? $\log _2\left (\frac {8}{16}\right ) = 3-\answer {4}$

What makes the following expression true? $\log _3(x) = \frac {\ln (x)}{\answer {\ln (3)}}$

Exponential equations

Let’s look into solving equations involving these functions. We’ll start with a straightforward example.

Solve the equation: $\displaystyle 4^x = 8$ .

We know $4$ and $8$ are each powers of $2$ , we start by rewriting in terms of this base. $4^x = 2^{2x} \,\,\, \textrm { and } \,\,\, 8 =2^{\answer {3}} \,\,\,\, \textrm { so } 2^{2x} = 2^{\answer {3}}$ Since exponential functions are one-to-one, the only way for $a^m = a^n$ is if $m=n$ . In this case, that means $\displaystyle 2x = 3$ .

The solution is: $\displaystyle x = \answer {3/2}$ .

Of course, if we couldn’t rewrite both sides with the same base, we can still use the properties of logarithms to solve.

Solve the equation: $\displaystyle 5^{2x-3} = 7$ .

Since we can’t easily rewrite both sides as exponentials with the same base, we’ll use logarithms instead. Above we said that $\log _b(x) = y$ means that $b^y = x$ . That statement means that each exponential equation has an equivalent logarithmic form and vice-versa. We’ll convert to a logarithmic equation and solve from there. $\begin{align*} 5^{2x-3} &= 7\\ \log_{\answer{5}}\left( \answer{7} \right) &= 2x-3 \end{align*}$

From here, we can solve for $x$ directly.

$\begin{align*} 2x &= \log_{5}\left(7\right) + 3\\ x &= \frac{\log_{5}\left(7\right) + 3}{2} \end{align*}$

Solve the equation: $\displaystyle e^{2x} = e^x + 6$ .

Immediately taking logarithms of both sides will not help here, as the right side has multiple terms. We know that logarithms do not behave well with sums, but with products/quotients. Instead, we notice that $e^{2x} = \left (e^x\right )^2$ . (This is a common trick that you will likely see many times.) $\begin{align*} e^{2x} &= e^x + 6\\ \left(e^x\right)^2 &= e^x + 6\\ \left(e^x\right)^2 - e^x - 6 &= 0 \end{align*}$

Our equation is really a quadratic equation in $e^x$ ! The left-hand side factors as $\left ( e^x - \answer {3}\right ) \left (e^x + \answer {2}\right )$ , so we are dealing with $e^x - \answer {3} = 0 \qquad \textrm {and} \qquad e^x+\answer {2} = 0.$ For the first:

$\begin{align*} e^x &= \answer{3}\\ x &= \ln\left( \answer{3}\right). \end{align*}$

From the second: $\displaystyle e^x = \answer {-2}$ . Look back at the graph of $y=e^x$ above. What was the range of the exponential function? It didn’t include any negative numbers, so $e^x = -2$ has no solutions.

The solution to $\displaystyle e^{2x} = e^x + 6$ is $x = \answer {\ln (3)}$ .

Solve the equation: $\displaystyle 2\left (5^{2x} + 6\right ) = 11 \cdot 5^x$ .

$\displaystyle \log _{5}\left (\dfrac {3}{2} \right )$ $\displaystyle \frac {\ln \left (\dfrac {3}{2} \right )}{5}$ $\displaystyle \log _{4}\left (5\right )$ $\displaystyle \log _{5}\left (4 \right )$ The equation has no solutions.

Solve the inequality: $\displaystyle \frac {6^x - 7 \cdot 3^x}{4^x - 15} \ge 0$ .

Since this isn’t a linear inequality, we’ll solve it using a sign-chart. Luckily, the right-side is already $0$ . Let’s factor the numerator on the left: $\begin{align*} 6^x - 7 \cdot 3^x &= \left(2\cdot 3\right)^x - 7 \cdot 3^x \\ &= 2^x \cdot 3^x - 7 \cdot 3^x\\ &= 3^x \left( 2^x - 7 \right). \end{align*}$

That means we need to construct a sign chart for $\displaystyle \frac {3^x \left ( 2^x - 7\right )}{4^x - 15}$ . (Note: $\log _{2}(7)$ is about $2.81$ and $\log _{4}(15)$ is about $1.95$ .)

The solution is: $\left ( -\infty , \, \log _{4}(15) \, \right ) \bigcup \left [ \,\log _{2}(7) \,, \infty \right )$

Logarithmic equations

Solve the equation: $\displaystyle \log _5( 2x+1) = 3$ .

Our first step will be to rewrite this logarithmic equation into its exponential form. $\log _5(2x+1) = 3 \qquad \textrm { means } \qquad 2x+1 = 5^{\answer {3}}$ From here we solve directly. $\begin{align*} 2x+1 &= \answer{125}\\ 2x &= \answer{124}\\ x &= \answer{62}. \end{align*}$

Solve the equation: $\log _3(2x+1) = 1-\log _3(x+2).$

With more than one logarithm, we’ll typically try to use the properties of logarithms to combine them into a single term. $\begin{align*} \log_3(2x+1) &= 1-\log_3(x+2) \\ \log_3(2x+1) + \log_3(x+2) &= 1\\ \log_3\left( (2x+1)(x+2)\right) &= 1\\ \log_3\left( 2x^2 + 5x + 2 \right) &= 1\\ 2x^2 + 5x + 2 &= 3\\ 2x^2 + 5x - 1 &= 0 \end{align*}$

Let’s use quadratic formula to solve this. $x = \frac {-5 \pm \sqrt {5^2 - 4 \cdot 2\cdot -1}}{2 \cdot 2} = \frac { -5\pm \sqrt { \answer {33} } }{4}.$

What happens if we try to plug $x = \dfrac {-5 - \sqrt {33}}{4}$ into the equation? Both $2\left ( \dfrac {-5-\sqrt {33}}{4} \right ) + 1$ and $\dfrac {-5-\sqrt {33}}{4} + 2$ are negative. That means, the logarithms of these values is not defined.

It turns out that $\dfrac {-5 - \sqrt {33}}{4}$ is a solution of the equation $2x^2+5x-1 = 0$ , but not a solution of the original equation $\log _3(2x+1) = 1-\log _3(x+2)$ .

When working with logarithmic equations, we must always check that the solutions we find actually satisfy the original equation.

The only solution is $x = \dfrac {-5 + \sqrt {33}}{4}$ .