SOLVING EXPONENTIAL EQUATIONS

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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:

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

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):

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

 

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.
  
  

  (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 $\,x\,,$ $\,b^x > 0\,.$

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 ($\,\frac 52 = 2.5\,$): $$ \begin{gather} \cssId{s119}{8\cdot 5^{2x-1} - 20 = 0}\cr\cr \cssId{s120}{8\cdot 5^{2x-1} = 20}\cr\cr \cssId{s121}{5^{2x-1} = \frac{5}{2}}\cr\cr \cssId{s122}{(2x-1)(\ln 5) = \ln 2.5}\cr\cr \cssId{s123}{2x\ln 5 - \ln 5 = \ln 2.5}\cr\cr \cssId{s124}{2x\ln 5 = \ln 2.5 + \ln 5}\cr\cr \cssId{s125}{x = \frac{\ln 12.5}{2\ln 5}} \end{gather} $$

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

 

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

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