Question

solve $(ln(x))^2 + 3ln(x) = -2$.
select one:
a. $x = e$ and $x = e^2$
b. $x = -1$ and $-2$
c. $x = e^{-1}$ and $x = e^{-2}$
d. $x = e^{-\frac{2}{5}}$

Explanation:

Step1: Substitute $u=\ln(x)$

Let $u = \ln(x)$. The equation becomes:
$$u^2 + 3u = -2$$

Step2: Rearrange to standard quadratic form

Rewrite into $ax^2+bx+c=0$:
$$u^2 + 3u + 2 = 0$$

Step3: Factor the quadratic equation

Factor the polynomial:
$$(u+1)(u+2) = 0$$

Step4: Solve for $u$

Set each factor equal to 0:
$u+1=0 \implies u=-1$; $u+2=0 \implies u=-2$

Step5: Substitute back $u=\ln(x)$

Solve for $x$ using $x=e^u$:
For $u=-1$: $x=e^{-1}$; For $u=-2$: $x=e^{-2}$

Answer:

C. $x = e^{-1}$ and $x = e^{-2}$