﻿ Solving Exponential Equations

# Solving Exponential Equations

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:    $\displaystyle \frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$
• Examplel 3:    $5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$

## The ‘IUSC’ Method

Here are the tools used in the IUSC method:

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 &= \frac{\ln 10}{3\ln 2}\end{align}

(divide by $\,3\ln 2\,$)

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

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

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

## Example 2

Solve:    $\displaystyle\frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$

$\displaystyle\frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$
Original Equation
$2^{3t-1} = 7\cdot 5^{2+t}$
Isolate

Multiply both sides by $\,7\cdot 5^{2+t}\,$ to get in the form ‘ExpTerm1 = ExpTerm2’. Note that $\,7\cdot 5^{2+t}\,$ is never equal to zero, so this equation is equivalent to the former.

$\ln (2^{3t-1}) = \ln (7\cdot 5^{2+t})$
Undo

Take natural logs of both sides. Note that both ‘$\, 2^{3t-1}\,$’ and ‘$\,7\cdot 5^{2+t}\,$’ are always positive, so this equation is equivalent to the former.

$(3t-1)(\ln 2) = \ln 7 + (2+t)(\ln 5)$
Undo (continued)

Use properties of logarithms to get the variables out of the exponents. This is now a linear equation in one variable.

$3t\ln 2 - t\ln 5 = \ln 7 + 2\ln 5 + \ln 2$
Solve

Get all the variable terms on the left, and constant terms on the right.

$t(3\ln 2 - \ln 5) = \ln 7 + 2\ln 5 + \ln 2$
Solve (continued)

Factor out the $\,t\,$ on the LHS.

$\displaystyle t = \frac{\ln 7 + 2\ln 5 + \ln 2}{3\ln 2 - \ln 5}$
Solve (continued)

Divide by $\,3\ln 2 - \ln 5\,.$

\displaystyle \begin{align} t &= \frac{\ln 7 + 2\ln 5 + \ln 2}{3\ln 2 - \ln 5}\cr\cr &\approx 12.46359 \end{align}

$\displaystyle\frac{2^{3(12.46359)-1}}{7\cdot 5^{2+12.46359}}\overset{\text{?}}{=} 1$

$1.00000 = 1$

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?

Here, the left-hand side evaluates to the number $\,1\,$ when rounded to five decimal places.

## Example 3

Solve:    $5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$

$5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$
Original Equation
$\ln \bigl(5^{x}\cdot 3^{x-1}\bigr) = \ln\bigl(\frac 12 7^{2-x}\bigr)$
Take logs.
$x\ln 5 + (x-1)\ln 3 = \ln(2^{-1}) + (2-x)\ln 7$
Use properties of logs.
$x(\ln 5 + \ln 3 + \ln 7) = -\ln 2 + \ln 3 + 2\ln 7$
Get variable terms on left and constants on right.
\displaystyle\begin{align} x &= \frac{-\ln 2 + \ln 3 + 2\ln 7}{(\ln 5 + \ln 3 + \ln 7)} \cr\cr &\approx 0.92336 \end{align}
Get an exact answer first, and then approximate.
$5^{0.92336}\cdot 3^{0.92336-1} \overset{\text{?}}{=} \frac 12 7^{2-0.92336}$

$4.06288 \approx 4.06290$

Okay!
Check in original equation. Approximate the solution to more than five decimal places, as needed and desired.

Solve:    ${\text{e}}^{2x} - {\text{e}}^x - 6 = 0$

This final exponential equation cannot be put in the form ‘ExpTerm1 = ExpTerm2’. However, it can be turned into a quadratic equation by a simple substitution. Consequently, it is often called a ‘fake quadratic’ or a ‘pseudo-quadratic’ equation.

The idea used is important:

• Have an equation you can't immediately solve?
• Try to transform it into one that you can solve!
• Solve the ‘transformed’ equation.
• Transform back to a solution of the original equation.
${\text{e}}^{2x} - {\text{e}}^x - 6 = 0$
Original Equation

This equation is not solvable by IUSC.

${(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0$

Use a property of exponents to rename the first term.

$u^2 - u - 6 = 0$

Let $\,u := {\text{e}}^x\,.$ This substitution transforms the equation in $\,x\,$ to a quadratic equation in $\,u\,.$

$(u-3)(u+2) = 0$

$u = 3$   or   $u = -2$

Solve the transformed equation. You can save a couple steps if you're comfortable never explicitly bringing $\,u\,$ into the picture:

$$\begin{gather} \cssId{sb74}{{(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0}\cr\cr \cssId{sb75}{({\text{e}}^{x} - 3)({\text{e}}^{x} + 2) = 0}\cr\cr \cssId{sb76}{{\text{e}}^x = 3\ \ \text{ or }\ \ {\text{e}}^x = -2} \end{gather}$$
${\text{e}}^x = 3$   or   ${\text{e}}^x = -2$

Transform back: go back to the original variable, $\,x\,.$

$x\ln \text{e} = \ln 3$

$x = \ln 3 \approx 1.09861$

Since $\,{\text{e}}^x\,$ is always strictly positive, it never equals a negative number. There is only one solution.

${\text{e}}^{2\,\cdot\, 1.09861} - {\text{e}}^{1.09861} - 6 \overset{\text{?}}{=} 0$

$-0.00003 \approx 0$

Okay!

Check.