Question

what is the true solution to $3\ln 2 + \ln 8 = 2\ln (4x)$?\
$x = 2$\
$x = 1$\
$x = 4$\
$x = 8$

Explanation:

Step1: Simplify left side via log rule

$3\ln 2 + \ln 8 = \ln 2^3 + \ln 8 = \ln 8 + \ln 8 = \ln(8 \times 8) = \ln 64$

Step2: Simplify right side via log rule

$2\ln(4x) = \ln(4x)^2 = \ln(16x^2)$

Step3: Equate arguments of logs

$\ln 64 = \ln(16x^2) \implies 64 = 16x^2$

Step4: Solve for $x^2$

$x^2 = \frac{64}{16} = 4$

Step5: Solve for positive $x$

$x = \sqrt{4} = 2$ (since $4x>0$, $x>0$)

Answer:

$x = 2$