what is the true solution to 3 ln 2+ln 8 = 2 ln(4x)?\no x = 1\no x = 2\no x = 4\no x = 8

Explanation:

Step1: Simplify the left - hand side using logarithm rules

Use the power rule $n\ln a=\ln(a^{n})$ and the product rule $\ln a+\ln b = \ln(ab)$. $3\ln 2+\ln 8=\ln(2^{3})+\ln 8=\ln 8+\ln 8=\ln(8\times8)=\ln 64$

Step2: Simplify the right - hand side

Using the power rule $n\ln a=\ln(a^{n})$, $2\ln(4x)=\ln((4x)^{2})=\ln(16x^{2})$

Step3: Set the two sides equal and solve for x

Since $\ln 64=\ln(16x^{2})$, then $64 = 16x^{2}$. Divide both sides by 16: $\frac{64}{16}=x^{2}$, so $x^{2}=4$. Taking the square root of both sides, we get $x = 2$ or $x=-2$. But since the argument of a logarithm $4x>0$ (for $\ln(4x)$ to be well - defined in the real number system), we discard $x = - 2$.

Answer:

$x = 2$