by Dr. Carol JVF Burns (website creator)
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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: $\,b > 0\,,$ $\,b\ne 1\,.$
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:

For example, these are all ExpTerms: These are NOT ExpTerms:

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


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


The ‘IUSC’ Method

Here are the tools used in the IUSC method:

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.



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.
$\displaystyle x = \frac{\ln 20 - \ln 8 + \ln 5}{2\ln 5} = \frac{\ln \frac{20\cdot 5}{8}}{2\ln 5} = \frac{\ln 12.5}{2\ln 5}$ Solve (continued):
Divide by $\,2\ln 5\,$; rename using properties of logs, if desired.
$\displaystyle x = \frac{\ln 12.5}{2\ln 5} \approx 0.78466$

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

$-0.00011 \approx 0$

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!

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
($\,\frac 52 = 2.5\,$):
$$ \begin{gather} \cssId{s119}{8\cdot 5^{2x-1} - 20 = 0}\cr\cr \cssId{s120}{8\cdot 5^{2x-1} = 20}\cr\cr \cssId{s121}{5^{2x-1} = \frac{5}{2}}\cr\cr \cssId{s122}{(2x-1)(\ln 5) = \ln 2.5}\cr\cr \cssId{s123}{2x\ln 5 - \ln 5 = \ln 2.5}\cr\cr \cssId{s124}{2x\ln 5 = \ln 2.5 + \ln 5}\cr\cr \cssId{s125}{x = \frac{\ln 12.5}{2\ln 5}} \end{gather} $$
$$ \begin{gather} \cssId{s126}{8\cdot 5^{2x-1} - 20 = 0}\cr\cr \cssId{s127}{8\cdot 5^{2x-1} = 20}\cr\cr \cssId{s128}{5^{2x-1} = \frac{5}{2}}\cr\cr \cssId{s129}{(2x-1)(\ln 5) = \ln 2.5}\cr\cr \cssId{s130}{2x - 1 = \frac{\ln 2.5}{\ln 5}}\cr\cr \cssId{s131}{2x = \frac{\ln 2.5}{\ln 5} + 1}\cr\cr \cssId{s132}{x = \frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)} \end{gather} $$ Note:
$\displaystyle \cssId{s134}{\frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)} \cssId{s135}{= \frac 12\left(\frac{\ln 2.5}{\ln 5} + \frac{\ln 5}{\ln 5}\right)} \cssId{s136}{= \frac 12\left(\frac{\ln 2.5 + \ln 5}{\ln 5}\right)} \cssId{s137}{= \frac{\ln 12.5}{2\ln 5}}$



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 $\,3\ln 2 - \ln 5\,.$
$\displaystyle t = \frac{\ln 7 + 2\ln 5 + \ln 2}{3\ln 2 - \ln 5} \approx 12.46359$

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

$1.00000 = 1$

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.


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.
$\displaystyle x = \frac{-\ln 2 + \ln 3 + 2\ln 7}{(\ln 5 + \ln 3 + \ln 7)} \approx 0.92336$ Get an exact answer first, and then approximate.
$5^{0.92336}\cdot 3^{0.92336-1} \overset{\text{?}}{=} \frac 12 7^{2-0.92336}$

$4.06288 \approx 4.06290$

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

EXAMPLE:   a ‘fake quadratic’

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:

${\text{e}}^{2x} - {\text{e}}^x - 6 = 0$ (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.
A familiar quadratic pattern emerges!
$u^2 - u - 6 = 0$ 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.
${\text{e}}^{2\,\cdot\, 1.09861} - {\text{e}}^{1.09861} - 6 \overset{\text{?}}{=} 0$

$-0.00003 \approx 0$

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Solving Logarithmic Equations
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
1 2 3 4 5 6

(MAX is 6; there are 6 different problem types.)