Solving Exponential Equations (Part 1)

(This page is Part 1. Click here for Part 2.)

An exponential equation has at least one variable in an exponent.

For example, ‘$\,2^{3x-1} = 5\,$’ is an exponential equation, since the variable $\,x\,$ is in the exponent.

However, ‘$(3x-1)^2 = 5$’ is not an exponential equation, since there is no variable in an exponent.

Many exponential equations can be solved by a technique that can be abbreviated as ‘IUSC’:

Isolate $\ldots$ Undo with a logarithm $\ldots$ Solve resulting equation $\ldots$ Check

This section classifies families of equations that can be solved by the IUSC method, discusses the method, and presents examples.

A technique for solving ‘fake’ quadratic (pseudo-quadratic) exponential equations is also presented.

Classification of Equations Solvable by IUSC

Let $\,b\,$ denote an allowable base for an exponential function: $\,b \gt 0\,,$ $\,b\ne 1\,.$

Recall that a ‘linear expression in one variable’ (say, $\,x\,$) is of the form $\,ax + b\,,$ for real numbers $\,a\,$ and $\,b\,.$

For the purposes of this section, define an ‘ExpTerm’ to be a single term that contains only the following types of factors:

constants
$b^{\text{linear expression in one variable}}$

For example, these are all ExpTerms:

$\,3\cdot 5^{2x-1}\,$
$\,7\,$
$\,2^x\cdot 7^{1-3x}\,$

These are not ExpTerms:

$\,2^x + 1\,$ (an ExpTerm is a single term)
$\,x3^x\,$ (the factor of $\,x\,$ is not allowed)

The IUSC method can be used to solve equations that can be put in the form:

ExpTerm1 = ExpTerm2

Here are examples of equations solvable by IUSC (and each is solved below):

Example 1: $8\cdot 5^{2x-1} - 20 = 0$
Example 2: $\displaystyle \frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$
Example 3: $5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$

The ‘IUSC’ Method

Here are the tools used in the IUSC method:

Solving Linear Equations in One Variable

The equations that emerge usually have irrational numbers involved (like ‘$\,\ln 3\,$’). Don't be intimidated by these ‘ugly’ numbers!

To prepare yourself, compare these side-by-side solutions of a ‘familiar’ equation and one you'll see in IUSC.

Familiar	IUSC
$2(3x-1) = 5$ (original equation)	$(3x-1)(\ln 2) = \ln 5$ (original equation)
$6x - 2 = 5$ (distributive law)	$3x(\ln 2) - \ln 2 = \ln 5$ (distributive law)
$6x = 7$ (isolate $\,x\,$ term)	$3x(\ln 2) = \ln 5 + \ln 2$ (isolate $\,x\,$ term)
$\displaystyle x = \frac{7}{6}$ (divide by $\,6\,$)	$\displaystyle \begin{align} x &= \frac{\ln 5 + \ln 2}{3\ln 2}\cr\cr &= \frac{\ln 10}{3\ln 2}\end{align}$ (divide by $\,3\ln 2\,$)

Properties of Logarithms

For all positive real numbers $\,x\,$ and $\,y\,,$ and for $\,s\in\Bbb R\,$:

$$ \begin{gather} x = y\ \ \iff\ \ \ln x = \ln y \cr\cr \ln x^s = s\ln x \cr \text{(you can bring exponents down)}\cr\cr \ln xy = \ln x + \ln y\cr \text{(the log of a product is the sum of the logs)}\cr\cr \ln\frac{x}{y} = \ln x - \ln y\cr \text{(the log of a quotient is the difference of the logs)} \end{gather} $$

Outputs From Exponential Functions are Always Positive

For all allowable bases $\,b\,$ and for all $\,x\,,$ $\,b^x \gt 0\,.$

The IUSC Method solving simple exponential equations

ISOLATE an exponential expression. That is, get an ExpTerm (which contains a variable) all by itself on one side of the equation.
UNDO the exponents with a logarithm. Any log can be used, but it's usually easiest to use the common log ($\,\log\,$) or the natural log ($\,\ln \,$).
SOLVE for the variable. Don't be intimidated by the irrational numbers! These are just linear equations in one variable.
CHECK in the original equation. There are lots of opportunities for errors in this method. To gain confidence, get a decimal approximation and substitute in the original equation.

Note: Always get an exact answer first. Then, get a decimal approximation (as needed) from the exact answer. Approximation errors made early on can grow as you proceed through the solution steps.

Example 1

Solve: $8\cdot 5^{2x-1} - 20 = 0$

$8\cdot 5^{2x-1} - 20 = 0$

Original Equation

$8\cdot 5^{2x-1} = 20$

Isolate

Add $\,20\,$ to both sides, to isolate an ExpTerm with a variable on the left. Some people prefer to divide through by $\,8\,$; this alternative solution is shown below.

$\ln\bigl(8\cdot 5^{2x-1}\bigr) = \ln 20$

Undo

Take natural logs of both sides. Note that both ‘$\,8\cdot 5^{2x-1}\,$’ and ‘$\,20\,$’ are always positive; therefore, this equation is equivalent to the first.

$\ln 8 + \ln 5^{2x-1} = \ln 20$

Undo (continued)

The log of a product is the sum of the logs.

$\ln 8 + (2x-1)(\ln 5) = \ln 20$

Undo (continued)

Bring the exponent down; this gets the variable out of the exponent. The result is a linear equation in one variable (involving several irrational numbers).

$(2x-1)(\ln 5) = \ln 20 - \ln 8$

Solve

Subtract $\,\ln 8\,$ from both sides.

$2x(\ln 5) - \ln 5 = \ln 20 - \ln 8$

Solve (continued)

Use the distributive law on the left side.

$2x(\ln 5) = \ln 20 - \ln 8 + \ln 5$

Solve (continued)

Add $\,\ln 5\,$ to both sides.

$\displaystyle \begin{align} x &= \frac{\ln 20 - \ln 8 + \ln 5}{2\ln 5}\cr\cr &= \frac{\ln \frac{20\cdot 5}{8}}{2\ln 5}\cr\cr &= \frac{\ln 12.5}{2\ln 5}\end{align}$

Solve (continued)

Divide by $\,2\ln 5\,$; rename using properties of logs, if desired.

$\displaystyle x = \frac{\ln 12.5}{2\ln 5} \approx 0.78466$

$8\cdot 5^{2(0.78466)-1} - 20 \overset{\text{?}}{=} 0$

$-0.00011 \approx 0$

Okay!

Check

Approximate the exact answer to at least five decimal places. Substitute in the original equation. Put a question mark over the equal sign, since you're asking a question—are the two sides equal?

Since you're using an approximate solution, you won't get a perfect equality, but it should be very close!

Here are two other approaches. Notice that different approaches can give different ‘names’ for the solution!

First Alternative Approach

By isolating $\,5^{2x-1}\,$ (instead of $\,8\cdot 5^{2x-1}\,$) before ‘undoing’ with logs, you must deal with a fraction, but save a couple steps overall. Here, we were lucky, because the fraction is an exact decimal ($\,\frac 52 = 2.5\,$):

$$ \begin{gather} \cssId{s119}{8\cdot 5^{2x-1} - 20 = 0}\cr\cr \cssId{s120}{8\cdot 5^{2x-1} = 20}\cr\cr \cssId{s121}{5^{2x-1} = \frac{5}{2}}\cr\cr \cssId{s122}{(2x-1)(\ln 5) = \ln 2.5}\cr\cr \cssId{s123}{2x\ln 5 - \ln 5 = \ln 2.5}\cr\cr \cssId{s124}{2x\ln 5 = \ln 2.5 + \ln 5}\cr\cr \cssId{s125}{x = \frac{\ln 12.5}{2\ln 5}} \end{gather} $$

Second Alternative Approach

$$ \begin{gather} \cssId{s126}{8\cdot 5^{2x-1} - 20 = 0}\cr\cr \cssId{s127}{8\cdot 5^{2x-1} = 20}\cr\cr \cssId{s128}{5^{2x-1} = \frac{5}{2}}\cr\cr \cssId{s129}{(2x-1)(\ln 5) = \ln 2.5}\cr\cr \cssId{s130}{2x - 1 = \frac{\ln 2.5}{\ln 5}}\cr\cr \cssId{s131}{2x = \frac{\ln 2.5}{\ln 5} + 1}\cr\cr \cssId{s132}{x = \frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)} \end{gather} $$

Note:

$$\begin{align} \cssId{s134}{\frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)} &\ \cssId{s135}{= \frac 12\left(\frac{\ln 2.5}{\ln 5} + \frac{\ln 5}{\ln 5}\right)}\cr\cr &\ \cssId{s136}{= \frac 12\left(\frac{\ln 2.5 + \ln 5}{\ln 5}\right)}\cr\cr &\ \cssId{s137}{= \frac{\ln 12.5}{2\ln 5}} \end{align} $$

Solving Exponential Equations (Part 1)

Classification of Equations Solvable by IUSC

The ‘IUSC’ Method

Example 1

First Alternative Approach

Second Alternative Approach

Concept Practice