An exponential equation has at least one variable in an exponent.
For example, ‘$\,2^{3x1} = 5\,$’ is an exponential equation, since the variable $\,x\,$ is in the exponent.
However, ‘$(3x1)^2 = 5$’ 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’:
A technique for solving ‘fake’ quadratic (pseudoquadratic) exponential equations is also presented.
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:
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):
Here are the tools used in the IUSC method:
(original equation)  $2(3x1) = 5$  $(3x1)(\ln 2) = \ln 5$  (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 $\,3\ln 2\,$) 
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: $8\cdot 5^{2x1}  20 = 0$
$8\cdot 5^{2x1}  20 = 0$  (original equation) 
$8\cdot 5^{2x1} = 20$ 
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. 
$\ln\bigl(8\cdot 5^{2x1}\bigr) = \ln 20$ 
Undo: Take natural logs of both sides. Note that both ‘$\,8\cdot 5^{2x1}\,$’ and ‘$\,20\,$’ are always positive; therefore, this equation is equivalent to the first. 
$\ln 8 + \ln 5^{2x1} = \ln 20$ 
Undo (continued): The log of a product is the sum of the logs. 
$\ln 8 + (2x1)(\ln 5) = \ln 20$ 
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). 
$(2x1)(\ln 5) = \ln 20  \ln 8$ 
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 $\,5^{2x1}\,$ (instead of
$\,8\cdot 5^{2x1}\,$) 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 ($\,\frac 52 = 2.5\,$): $$ \begin{gather} \cssId{s119}{8\cdot 5^{2x1}  20 = 0}\cr\cr \cssId{s120}{8\cdot 5^{2x1} = 20}\cr\cr \cssId{s121}{5^{2x1} = \frac{5}{2}}\cr\cr \cssId{s122}{(2x1)(\ln 5) = \ln 2.5}\cr\cr \cssId{s123}{2x\ln 5  \ln 5 = \ln 2.5}\cr\cr \cssId{s124}{2x\ln 5 = \ln 2.5 + \ln 5}\cr\cr \cssId{s125}{x = \frac{\ln 12.5}{2\ln 5}} \end{gather} $$ 
$$
\begin{gather}
\cssId{s126}{8\cdot 5^{2x1}  20 = 0}\cr\cr
\cssId{s127}{8\cdot 5^{2x1} = 20}\cr\cr
\cssId{s128}{5^{2x1} = \frac{5}{2}}\cr\cr
\cssId{s129}{(2x1)(\ln 5) = \ln 2.5}\cr\cr
\cssId{s130}{2x  1 = \frac{\ln 2.5}{\ln 5}}\cr\cr
\cssId{s131}{2x = \frac{\ln 2.5}{\ln 5} + 1}\cr\cr
\cssId{s132}{x = \frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)}
\end{gather}
$$
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: $\displaystyle\frac{2^{3t1}}{7\cdot 5^{2+t}} = 1$
$\displaystyle\frac{2^{3t1}}{7\cdot 5^{2+t}} = 1$  (original equation) 
$2^{3t1} = 7\cdot 5^{2+t}$ 
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. 
$\ln (2^{3t1}) = \ln (7\cdot 5^{2+t})$ 
Undo: Take natural logs of both sides. Note that both ‘$\, 2^{3t1}\,$’ and ‘$\,7\cdot 5^{2+t}\,$’ are always positive, so this equation is equivalent to the former. 
$(3t1)(\ln 2) = \ln 7 + (2+t)(\ln 5)$ 
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 lefthand side evaluates to the number $\,1\,$ when rounded to five decimal places. 
$5^{x}\cdot 3^{x1} = \frac 12 7^{2x}$  (original equation) 
$\ln \bigl(5^{x}\cdot 3^{x1}\bigr) = \ln\bigl(\frac 12 7^{2x}\bigr)$  Take logs. 
$x\ln 5 + (x1)\ln 3 = \ln(2^{1}) + (2x)\ln 7$  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. 
$5^{0.92336}\cdot 3^{0.923361} \overset{\text{?}}{=} \frac 12 7^{20.92336}$ $4.06288 \approx 4.06290$ 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 ‘pseudoquadratic’ 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 $\,u\,$. 
$(u3)(u+2) = 0$ $u = 3$ 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. 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
