Question

If $3^{4x}=(81{)}^{-1}$ and $(10{)}^y=0.0001$, find the value of $2^{-x}\times {16}^y$

Answer 1

$3^{4x}=(81{)}^{-1}$

$3^{4x}=(3^4{)}^{-1}$

$3^{4x}=3^{-4}$

Base is same the powers can be equated

$\therefore 4x=-4⟹x=-1$

${10}^y=0.0001$

${10}^y={10}^{-4}$

Base is same the powers can be equated

$\therefore y=-4$

The value of $2^{-x}\times {16}^y$ is

$=2^{-(-1)}\times {16}^{-4}$

$=2\times \frac{1}{65536}$

$=\frac{1}{32768}$