If $ a $ and $ b $ are distinct positive primes such that $ \sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2 y} $, find $ x $ and $ y $.

Given:

\( a \) and \( b \) are distinct positive primes such that \( \sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2 y} \).

To do: 

We have to find \( x \) and \( y \).

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,

$\sqrt[3]{a^{6} b^{-4}}=a^{x} b^{2 y}$

$(a^{6} b^{-4})^{\frac{1}{3}}=a^{x} b^{2 y}$

$a^{\frac{6}{3}} \times b^{\frac{-4}{3}}=a^{x} \times b^{2 y}$

$a^{2} \times b^{\frac{-4}{3}}=a^{x} \times b^{2 y}$

Comparing both sides, we get,

$a^{x}=a^{2}$

$\Rightarrow x=2$

$b^{\frac{-4}{3}}=b^{2 y}$

$\Rightarrow 2y=\frac{-4}{3}$

$\Rightarrow y=\frac{-4}{3 \times 2}=\frac{-2}{3}$

The values of $x$ and $y$ are $2$ and $\frac{-2}{3}$ respectively.  

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Updated on: 10-Oct-2022

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