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$$ x = \left(\frac{4^y}{-2}\right)^{\frac{1}{3}} $$ 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.

derpy
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user118512
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  • original question has -2. you are correct and this is why i knew i was on the wrong path, but not sure how else to proceed – user118512 Apr 15 '14 at 16:48
  • Is that definitely the correct answer. Since if you put $y=2$ into the top formula you get $x=2(-1)^\frac{1}{3}$ but if you then put that into your answer you get: $y=16+\ln(4) \neq 2$. – Jay Apr 15 '14 at 16:59
  • yes, book answer is above. i was also not sure how they picked base 4 except it was in the numerator of the original. – user118512 Apr 15 '14 at 17:06
  • You could solve it using base-4 by writing as: $-2x^3=4^y$ then taking log to the base-4 of both sides giving: $\log_4(-2x^3)=y$. But that still the same answer as the one in my answer since: $\log_a(b)=\frac{\ln(b)}{\ln{(a)}}$. – Jay Apr 15 '14 at 17:14
  • how are you getting rid of the (1/3) exponent to make it look like you did? – user118512 Apr 15 '14 at 17:52
  • cubed both sides, so $(x)^3=\left(\left(\frac{4^y}{-2}\right)^\frac{1}{3}\right)^3$ which is the same as: $ x^3= \left(\frac{4^y}{-2}\right)^1=\frac{4^y}{-2}$ then multiply both sides by $-2$ to get: $-2x^3=4^y$. – Jay Apr 15 '14 at 17:59
  • thanks bud, i see it clearly now! i was trying to follow my log do-undo steps but i failed to think about raising a power to a power. thanks for your time – user118512 Apr 15 '14 at 18:02
  • No worries, happy to help! – Jay Apr 15 '14 at 18:06

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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: $$ \ln(x)=\frac{1}{3}\ln \left(\frac{4^y}{-2}\right) \\ \ln(x^3)= \ln \left(\frac{4^y}{-2}\right)\\ x^3=\frac{4^y}{-2} \\ -2x^3=4^y \\ \ln(-2x^3)=y\ln(4)\\ \therefore \;\; y=\frac{\ln(-2x^3)}{\ln(4)} $$

Jay
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