Question

solve: $e^{2x} - 7e^{x} = 18$

a. $x = \ln 18$

b. $x = 9$ and $x = -2$

c. $x = \ln 9$

d. $x = \ln 2$

Explanation:

Step1: Substitute $u=e^x$

$u^2 - 7u = 18$

Step2: Rearrange to standard quadratic

$u^2 - 7u - 18 = 0$

Step3: Factor the quadratic

$(u - 9)(u + 2) = 0$

Step4: Solve for $u$

$u = 9$ or $u = -2$

Step5: Substitute back $u=e^x$, discard invalid solution

$e^x = 9$ (since $e^x>0$, $u=-2$ is invalid)

Step6: Solve for $x$

$x = \ln 9$

Answer:

C. $x = \ln 9$