Find: \(2^{1+x}\), if \(3^{4x} = 81^{-1}\) and \(10^{1/y} = 0.0001\)
Solution
\[ 3^{4x} = 81^{-1} \]
\[ \Rightarrow 3^{4x} = (3^4)^{-1} \]
\[ \Rightarrow 3^{4x} = 3^{-4} \]
\[ \Rightarrow 4x = -4 \]
\[ \Rightarrow x = -1 \]
\[ 10^{1/y} = 0.0001 \]
\[ \Rightarrow 10^{1/y} = 10^{-4} \]
\[ \Rightarrow \frac{1}{y} = -4 \]
\[ \Rightarrow y = -\frac{1}{4} \]
\[ 2^{1+x} = 2^{1-1} \]
\[ = 2^0 \]
\[ = 1 \]
Final Answer:
\[ \boxed{1} \]