Solving Exponential Equations (Part 2)
(This page is Part 2. Click here for Part 1.)
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:
$$ \begin{gather} \cssId{s74}{{(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0}\cr\cr \cssId{s75}{({\text{e}}^{x} - 3)({\text{e}}^{x} + 2) = 0}\cr\cr \cssId{s76}{{\text{e}}^x = 3\ \ \text{ or }\ \ {\text{e}}^x = -2} \end{gather} $$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.