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’:
A technique for solving ‘fake’ quadratic (pseudo-quadratic) 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) | $2(3x-1) = 5$ | $(3x-1)(\ln 2) = \ln 5$ | (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 $\,3\ln 2\,$) |
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. |
$\displaystyle x = \frac{\ln 12.5}{2\ln 5} \approx 0.78466$ $8\cdot 5^{2(0.78466)-1} - 20 \overset{\text{?}}{=} 0$ $-0.00011 \approx 0$ 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: $\displaystyle \cssId{s134}{\frac 12\left(\frac{\ln 2.5}{\ln 5} + 1\right)} \cssId{s135}{= \frac 12\left(\frac{\ln 2.5}{\ln 5} + \frac{\ln 5}{\ln 5}\right)} \cssId{s136}{= \frac 12\left(\frac{\ln 2.5 + \ln 5}{\ln 5}\right)} \cssId{s137}{= \frac{\ln 12.5}{2\ln 5}}$ |
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. |
$5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$ | (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. |
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. |
${\text{e}}^{2\,\cdot\, 1.09861} - {\text{e}}^{1.09861} - 6 \overset{\text{?}}{=} 0$ $-0.00003 \approx 0$ 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:
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