Exponential Equations and Inequalities

In this page, we discuss how to solve equations and inequalities involving exponential functions. We begin with some basic cases and quickly evolve to more general situations.

You have already seen some of these cases in S5, when you discussed exponential models. The trick to solving these equations is to move them from an exponential equation into an equation that we can solve more easily.

Table of Contents

Basis Equations from S5

There are some cases that can be dealt with rather easily and intuitively:

Case 1: An exponential equation of the form $a^{f(x)}=a^{g(x)}$ is equivalent to $f(x)=g(x)$ .
Case 2: An exponential equation of the form $a^{f(x)}=b^{f(x)}$ is equivalent to $f(x)=0$ .

Let us see examples of this.

Example: Solve the exponential equation $5^{x^2 - 2x - 5} = 5^{- x - 3}$ for all real values of $x$ .

This was our typical easy example. If the bases are the same, the powers must be the same. See how we used that fact to go from an exponential equation to a quadratic one. The next example shows how to deal with the same power but different bases.

Example: Solve the exponential equation $3^{x^2 - x - 6} = 7^{x^2 - x - 6}$ for all real values of $x$ .

Our final basic example will deal with a case that is very similar to our first case. This is the case where the bases are different, but easily linked. Using properties of powers, we can turn these equations into the case where the bases are the same.

Example: Solve the exponential equation $4^{x^2 - x - 2} = 16^{x + 1}$ for all real values of $x$ .

Exponential Equations Resulting in Logarithms

So far, we have seen different combinations of situations: bases and powers being the same, different, or linked in some form. How do we deal with a situation that does not fall into this scenario?

For example, how do we solve $3^{x+2}=7^{x-1}$ ? The numbers 3 and 7 are not “easily linked” in some form using properties of powers. Here we use the logarithm. We can choose any number for the base, but some choices are clearly better than others.

Example: Solve the exponential equation $3^{x+2} = 7^{x-1}$ using three different logarithmic bases: base-3 ( $\log_3$ ), base-7 ( $\log_7$ ), and the natural logarithm ( $\ln$ ).

We can also approximate $\log_7(3),\log_3(7),\ln(7)$ , and $\ln(3)$ in the middle of our solutions and then approximate the answer. None of this changes the actual logic. Ideally, you take these elements as they are (fixed numbers) and only approximate your final answer.

Exponential Inequalities

Inequalities are in general slightly more annoying, but there is a general logic that can simplify some of them. If you want to understand where a function $f(x)$ is positive, find out where $f$ is 0 or undefined, then use a sign diagram and check between each of these possible places. This uses, again, our intuition that a function cannot change signs without having a point of discontinuity or a 0. Of course, an exponential function $a^{f(x)}$ is never negative since we only deal with $a>0$ so the inequalities will not be as simple. Instead, we compare two exponential functions.

Comparing two exponential functions in an inequality can be somewhat easier. If the basis is more than 1, then we move the inequality to the powers:

$a^{f(x)}>a^{g(x)}\iff f(x)>g(x)\text{ if }a>0$

If the basis is between 0 and 1, that is $0<a<1$ , then we swap the inequality sign.

$a^{f(x)}>a^{g(x)}\iff f(x)<g(x)$

Using these two principles, we can solve any number of simple exponential inequalities, albeit with some algebraic manipulations first.

Example: Solve the exponential inequality $\left(\frac{1}{3}\right)^{2x - 5} \le 9^{x + 2}$ for all real values of $x$ . State the final answer using interval notation.

We can also use a graphic calculator to observe this. Let us see an example with a quadratic function in the power:

Example: Solve the exponential inequality $\left(\frac{1}{2}\right)^{x^2 - 2x} > \left(\frac{1}{2}\right)^{x + 4}$ for all real values of $x$ . State the final solution using interval notation.

Exercises

For each of the following exponential equations, solve for all real values of $x$ . State your answers as exact values, and simplify any fractions or radicals where applicable.

$3^{x^2 - 4x - 7} = 3^{x + 7}$

$7^{2x^2 - x - 1} = 7^{x^2 + 5x + 6}$

$2^{x^2 - 3x} = 4^{x + 3}$

$9^{x^2 - x - 4} = 27^{x + 2}$

$5^{x^2 - 5x + 3} = \left(\frac{1}{25}\right)^{x - 1}$

$4^{x^2 - 2x - 8} = 9^{x^2 - 2x - 8}$

$2^{2x^2 - 7x + 3} = 5^{2x^2 - 7x + 3}$

$\left(\frac{2}{3}\right)^{x^2 - 9} = \left(\frac{5}{8}\right)^{x^2 - 9}$

For each of the following exponential equations, the bases cannot be converted into a common integer power. Solve for all real values of $x$ using logarithms. State your final answer as a quotient using either natural logarithms ( $\ln$ ) or any other logarithm, then approximate it.

$2^{x+3} = 5^{x-1}$

$3^{2x-1} = 4^{x+2}$

$6^{x+1} = 5^{2x-3}$

$10^{x-2} = 3^{3x+1}$

For each of the following exponential inequalities, find all real values of $x$ that satisfy the statement. Pay close attention to whether the base is greater than 1 or between 0 and 1, as this affects the inequality direction. State your final solution using interval notation.

$4^{3x - 7} > 4^{x + 5}$

$\left(\frac{1}{5}\right)^{2x + 9} \le \left(\frac{1}{5}\right)^{x - 3}$

$27^{x - 1} \ge \left(\frac{1}{9}\right)^{2x + 4}$

$3^{x^2 - 5x} < 3^{-6}$

$\left(\frac{1}{3}\right)^{x^2 - 3} \ge \left(\frac{1}{3}\right)^{2x}$

$2^{x^2 - 4x - 2} \le 8^{x + 2}$

Consider the exponential inequality:

$5^{\cos(x)} \le 5^{\sin(x)}$

Solve for all real values of $x$ restricted to the bounded interval $x \in [0, 2\pi)$ .