Question

solve $e^{2x} - 8e^{x} = 20$

select one:
a. $x = \ln(20)$
b. $x = \ln(2)$
c. $x = \ln(10)$
d. $x = 10$ and $x = -2$

Explanation:

Step1: Substitute $y=e^x$

Let $y = e^x$, then the equation becomes:
$$y^2 - 8y = 20$$

Step2: Rearrange to standard quadratic form

$$y^2 - 8y - 20 = 0$$

Step3: Factor the quadratic

$$(y - 10)(y + 2) = 0$$

Step4: Solve for $y$

Set each factor equal to 0:
$y - 10 = 0 \implies y = 10$; $y + 2 = 0 \implies y = -2$

Step5: Substitute back $y=e^x$

Since $e^x > 0$ for all real $x$, $y=-2$ has no solution. For $y=10$:
$$e^x = 10$$
Take natural log of both sides:
$$x = \ln(10)$$

Answer:

C. $x = \ln(10)$