Log Equations

$\log_{x}{2y}=3\color{#D61F06}{\Leftrightarrow}\color{#333333}x^{3}=2y$ $\log_{x}{4y}=2\color{#D61F06}{\Leftrightarrow}\color{#333333}x^{2}=4y$ It's a little hard to see at first, but substitute the second equation into the first one: $x^{3}=2y\color{#D61F06}{\Rightarrow}\color{#333333}x\cdot x^{2}=2y\color{#D61F06}{\Rightarrow}\color{#333333}x\cdot 4y=2y\color{#D61F06}{\Rightarrow}\color{#333333}x=\frac{1}{2}$

Note: That last step implies $y\neq 0$ . Since you can't take the logarithm of $0$ , this implication is fine to make.

Now it's simple to find y: $(\frac{1}{2})^{2}=4y\color{#D61F06}{\Rightarrow}\color{#333333}y=\frac{1}{16}$ And now to answer the question: $y^{-x}=(\frac{1}{16})^{-\frac{1}{2}}=\color{#3D99F6}{4}\space\space\space\Box$

Subtract the second equation from the first to obtain $\log_x \dfrac{1}{2} = 1 \rightarrow x = \dfrac{1}{2}$ .

Then we have by the second equation $x^2 = \dfrac{1}{4} = 4y \rightarrow y = \dfrac{1}{16}.$

Thus, $y^{-x} = \Big(\dfrac{1}{16}\Big)^{-1/2} = 4.$

Raise $x$ to the power of both sides $2y = x^3 \\ 4y = x^2$ and then divide one equation by the other, and the solution falls into place easily.