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:
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 $\,6\,$) |
$\displaystyle \begin{align} x &= \frac{\ln 5 + \ln 2}{3\ln 2}\cr &= \frac{\ln 10}{3\ln 2}\end{align}$ (divide by $\,3\ln 2\,$) |
For all positive real numbers $\,x\,$ and $\,y\,,$ and for $\,s\in\Bbb R\,$:
Outputs From Exponential Functions are Always Positive
For all allowable bases $\,b\,$ and for all $\,x\,,$ $\,b^x \gt 0\,.$
- 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: $8\cdot 5^{2x-1} - 20 = 0$
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.
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.
The log of a product is the sum of the logs.
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).
Subtract $\,\ln 8\,$ from both sides.
Use the distributive law on the left side.
Add $\,\ln 5\,$ to both sides.
Divide by $\,2\ln 5\,$; rename using properties of logs, if desired.
$8\cdot 5^{2(0.78466)-1} - 20 \overset{\text{?}}{=} 0$
$-0.00011 \approx 0$
Okay!
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:
$$\begin{align} \cssId{s134}{\frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)}\ &\ \cssId{s135}{= \frac 12\left(\frac{\ln 2.5}{\ln 5} + \frac{\ln 5}{\ln 5}\right)}\cr &\ \cssId{s136}{= \frac 12\left(\frac{\ln 2.5 + \ln 5}{\ln 5}\right)}\cr &\ \cssId{s137}{= \frac{\ln 12.5}{2\ln 5}} \end{align} $$Read-Through, Part 2
Example 2
Solve: $\displaystyle\frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$
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.
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.
Use properties of logarithms to get the variables out of the exponents. This is now a linear equation in one variable.
Get all the variable terms on the left, and constant terms on the right.
Factor out the $\,t\,$ on the LHS.
Divide by $\,3\ln 2 - \ln 5\,.$
$\displaystyle\frac{2^{3(12.46359)-1}}{7\cdot 5^{2+12.46359}}\overset{\text{?}}{=} 1$
$1.00000 = 1$
Okay!
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: $5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$
$4.06288 \approx 4.06290$
Okay!
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.
This equation is not solvable by IUSC.
Use a property of exponents to rename the first term.
A familiar quadratic pattern emerges!
Let $\,u := {\text{e}}^x\,.$ This substitution transforms the equation in $\,x\,$ to a quadratic equation in $\,u\,.$
$u = 3$ or $u = -2$
Solve the transformed equation. You can save a couple steps if you're comfortable never explicitly bringing $\,u\,$ into the picture:
$$ \begin{gather} \cssId{sb74}{{(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0}\cr\cr \cssId{sb75}{({\text{e}}^{x} - 3)({\text{e}}^{x} + 2) = 0}\cr\cr \cssId{sb76}{{\text{e}}^x = 3\ \ \text{ or }\ \ {\text{e}}^x = -2} \end{gather} $$Transform back: go back to the original variable, $\,x\,.$
$x = \ln 3 \approx 1.09861$
Since $\,{\text{e}}^x\,$ is always strictly positive, it never equals a negative number. There is only one solution.
$-0.00003 \approx 0$
Okay!
Check.