problem: If $\displaystyle \left(\frac{3}{4}\right)^x=\frac{81}{256}$, what must $x$ be?
solution: Remembering that $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$, we have $$ \left(\frac{3}{4}\right)^x=\frac{3^x}{4^x}=\frac{81}{256}$$Comparing numerators, $3^x=81$ so $x=4$.
  
  Indeed, for the denominators we have $4^x=4^4=256$, as desired. Thus, $x=\boxed{4}$.