Find $x$ such that $\log_x 4=\log_{27} 3$.
I notice that the bases of the logarithms are different, so I want to convert them to a common base. I also notice that 27 is a power of 3, so I can use the change of base formula to write $\log_{27} 3$ as $\frac{\log_3 3}{\log_3 27}$. Since $\log_3 3=1$ and $\log_3 27=3$, this simplifies to $\frac{1}{3}$. So I have $\log_x 4=\frac{1}{3}$. To get rid of the logarithm, I can exponentiate both sides with base x. This gives me $x^{\log_x 4}=x^{\frac{1}{3}}$. By the definition of logarithm, $x^{\log_x 4}=4$, so I have $4=x^{\frac{1}{3}}$. To isolate x, I can cube both sides. This gives me $64=x$. I can check my answer by plugging it back into the original equation and verifying that it is true. 