# SOLVING EXPONENTIAL EQUATIONS

• PRACTICE (online exercises and printable worksheets)

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:

• constants
• $b^{\text{linear expression in one variable}}$
For example, these are all ExpTerms:
• $\,3\cdot 5^{2x-1}\,$
• $\,7\,$
• $\,2^x\cdot 7^{1-3x}\,$
There 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):

## 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.

 (original equation) $2(3x-1) = 5$ $(3x-1)(\ln 2) = \ln 5$ (original equation) (distributive law) $6x - 2 = 5$ $3x(\ln 2) - \ln 2 = \ln 5$ (distributive law) (isolate $\,x\,$ term) $6x = 7$ $3x(\ln 2) = \ln 5 + \ln 2$ (isolate $\,x\,$ term) (divide by $\,6\,$) $\displaystyle x = \frac{7}{6}$ $\displaystyle x = \frac{\ln 5 + \ln 2}{3\ln 2} = \frac{\ln 10}{3\ln 2}$ (divide by $\,3\ln 2\,$)
• properties of logarithms
For all positive real numbers $\,x\,$ and $\,y\,$, and for $\,s\in\Bbb R\,$: \begin{alignat}{2} x = y\ \ &\iff\ \ \ln x = \ln y &&\cr\cr \ln x^s &\, = s\ln x &&\text{(you can bring exponents down)}\cr\cr \ln xy &\, = \ln x + \ln y &\qquad &\text{(the log of a product is the sum of the logs)}\cr\cr \ln\frac{x}{y} &\, = \ln x - \ln y &&\text{(the log of a quotient is the difference of the logs)} \end{alignat}
• outputs from exponential functions are always positive
For all allowable bases $\,b\,$ and for all $\,x\,$, $\,b^x > 0\,$.
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

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. $\displaystyle x = \frac{\ln 12.5}{2\ln 5} \approx 0.78466$ $8\cdot 5^{2(0.78466)-1} - 20 \overset{\text{?}}{=} 0$ $-0.00011 \approx 0$ 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!

 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 ($\,\frac 52 = 2.5\,$): $$\begin{gather} 8\cdot 5^{2x-1} - 20 = 0\cr\cr 8\cdot 5^{2x-1} = 20\cr\cr 5^{2x-1} = \frac{5}{2}\cr\cr (2x-1)(\ln 5) = \ln 2.5\cr\cr 2x\ln 5 - \ln 5 = \ln 2.5\cr\cr 2x\ln 5 = \ln 2.5 + \ln 5\cr\cr x = \frac{\ln 12.5}{2\ln 5} \end{gather}$$ $$\begin{gather} 8\cdot 5^{2x-1} - 20 = 0\cr\cr 8\cdot 5^{2x-1} = 20\cr\cr 5^{2x-1} = \frac{5}{2}\cr\cr (2x-1)(\ln 5) = \ln 2.5\cr\cr 2x - 1 = \frac{\ln 2.5}{\ln 5}\cr\cr 2x = \frac{\ln 2.5}{\ln 5} + 1\cr\cr x = \frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right) \end{gather}$$ Note: $\displaystyle \frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right) = \frac 12\left(\frac{\ln 2.5}{\ln 5} + \frac{\ln 5}{\ln 5}\right) = \frac 12\left(\frac{\ln 2.5 + \ln 5}{\ln 5}\right) = \frac{\ln 12.5}{2\ln 5}$

## EXAMPLE

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 (continued): 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 $\,3\ln 2 - \ln 5\,$. $\displaystyle t = \frac{\ln 7 + 2\ln 5 + \ln 2}{3\ln 2 - \ln 5} \approx 12.46359$ $\displaystyle\frac{2^{3(12.46359)-1}}{7\cdot 5^{2+12.46359}}\overset{\text{?}}{=} 1$ $1.00000 = 1$ 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

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. $5^{0.92336}\cdot 3^{0.92336-1} \overset{\text{?}}{=} \frac 12 7^{2-0.92336}$ $4.06288 \approx 4.06290$ Okay! 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:

• 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. 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: $$\begin{gather} {(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0\cr\cr ({\text{e}}^{x} - 3)({\text{e}}^{x} + 2) = 0\cr\cr {\text{e}}^x = 3\ \ \text{ or }\ \ {\text{e}}^x = -2 \end{gather}$$ ${\text{e}}^x = 3$   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.
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.
PROBLEM TYPES:
 1 2 3 4 5 6
AVAILABLE MASTERED IN PROGRESS
 (MAX is 6; there are 6 different problem types.)