Finding inverse using logs

Question

$$ x = \left(\frac{4^y}{-2}\right)^{\frac{1}{3}} $$ i have correct answer of $\:y=\log(4)-2x^3$ i'm lost on steps to obtain the answer. i tried the following:$$\log\:x=\log\left(\frac{4^y}{-2}\right)^{\left(\frac{1}{3}\right)}\implies\log\:x=\left(\frac{1}{3}\right)\log\left(\frac{4^y}{-2}\right)\implies \log\:x=\left(\frac{1}{3}\right)\left(\log 4^y-\log \left(-2\right)\right)$$ but was lost at that point.

Answer 1

If your trying to solve for $x$ in the equation: $$ x=\left( \frac{4^y}{-2}\right)^{\frac{1}{3}} $$ Then you could manipulate it to solve for $y$ but I'm not entirely sure this is accurate given the negative logarithms: $$ \ln(x)=\frac{1}{3}\ln \left(\frac{4^y}{-2}\right) \\ \ln(x^3)= \ln \left(\frac{4^y}{-2}\right)\\ x^3=\frac{4^y}{-2} \\ -2x^3=4^y \\ \ln(-2x^3)=y\ln(4)\\ \therefore \;\; y=\frac{\ln(-2x^3)}{\ln(4)} $$