Question

If $3^{4x}={\left(81\right)}^{-1}$ and ${10}^{\frac{1}{y}}=0.0001$, find the value of $2^{-x+4y}$.

Answer 1

It is given that $3^{4x}={\left(81\right)}^{-1}$ and ${10}^{\frac{1}{y}}=0.0001$.
Now,
$$\begin{gathered} 3^{4x}={\left(81\right)}^{-1} \\ \Rightarrow 3^{4x}={\left(3^4\right)}^{-1} \\ \Rightarrow {\left(3^x\right)}^4={\left(3^{-1}\right)}^4 \\ \Rightarrow x=-1 \end{gathered}$$
And,
$$\begin{gathered} {10}^{\frac{1}{y}}=0.0001 \\ \Rightarrow {10}^{\frac{1}{y}}=\frac{1}{10000} \\ \Rightarrow {10}^{\frac{1}{y}}={\left(\frac{1}{10}\right)}^4 \\ \Rightarrow {10}^{\frac{1}{y}}={\left(10\right)}^{-4} \\ \Rightarrow \frac{1}{y}=-4 \\ \Rightarrow y=-\frac{1}{4} \end{gathered}$$
Therefore, the value of $2^{-x+4y}$ is $2^{1+4\left(-\frac{1}{4}\right)}=2^0=1$.