Question

ln(x² - 1) + 2ln2 = ln(4x - 1)

Explanation:

Step1: Simplify left side using log rules

$2\ln2 = \ln2^2 = \ln4$, so left side is $\ln(x^2-1) + \ln4 = \ln[4(x^2-1)]$

Step2: Set arguments equal (log equality)

$\ln[4(x^2-1)] = \ln(4x-1) \implies 4(x^2-1) = 4x-1$

Step3: Expand and rearrange to quadratic

$4x^2 - 4 = 4x -1 \implies 4x^2 -4x -3 =0$

Step4: Solve quadratic equation

Factor: $(2x-3)(2x+1)=0 \implies x=\frac{3}{2}$ or $x=-\frac{1}{2}$

Step5: Check domain validity

$x^2-1>0 \implies |x|>1$; $4x-1>0 \implies x>\frac{1}{4}$. So $x=-\frac{1}{2}$ is invalid.

Answer:

$\frac{3}{2}$