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

Given:

$3^{4x} = (81)^{-1}$ and $10^{\frac{1}{y}} = 0.0001$

To do: 

We have to find the value of $2^{-x+4y}$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

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

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

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

Comparing both sides, we get,

$4 x=-4$

$\Rightarrow x=\frac{-4}{4}$

$\Rightarrow x=-1$ $10^{\frac{1}{y}}=0.0001$

$=\frac{1}{10000}$

$=\frac{1}{(10)^{4}}$

$=(10)^{-4}$

Comparing both sides, we get,

$\frac{1}{y}=-4$

$\Rightarrow y=-\frac{1}{4}$

Therefore,

$2^{-x+4 y}=2^{-(-1)+4(-\frac{1}{4}}$

$=2^{1-1}$

$=2^{0}$

$=1$

The value of $2^{-x+4y}$ is $1$.  

Tutorialspoint

Updated on: 10-Oct-2022

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