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Exponents Of Real Numbers — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsExponents Of Real Numbers3 Marks
Question
If $a$ and $b$ are different positive primes such that $\Big(\frac{\text{a}^{-1}\text{b}^2}{\text{a}^2\text{b}^2}\Big)^7\div\Big(\frac{\text{a}^3\text{b}^{-5}}{\text{a}^{-1}\text{b}^3}\Big)=\text{a}^\text{x}\text{b}^\text{y},$ find $x$ and $y$.
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Answer

$\Big(\frac{\text{a}^{-1}\text{b}^2}{\text{a}^2\text{b}^2}\Big)^7\div\Big(\frac{\text{a}^3\text{b}^{-5}}{\text{a}^{-1}\text{b}^3}\Big)=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(​\text{a}^{-1-2}\times\text{b}^{2+4})^7\div(\text{a}^{3+2}\times\text{b}^{-5-3})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{a}^{-3}\times\text{b}^{6})\div(\text{a}^5\times\text{b}^{-\text{8}})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow(\text{a}^{-21}\times\text{b}^{42})\div(\text{a}^5\times\text{b}^{-8})=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\frac{\text{a}^{21}\times\text{b}^{42}}{\text{a}^5\times\text{b}^{-8}}=\text{a}^\text{x}\text{b}^\text{y}$
$\Rightarrow\text{a}^{-21-5}\times\text{b}^{42+8}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{a}^{-26}\times\text{b}^{50}=\text{a}^\text{x}\times\text{b}^\text{y}$
$\Rightarrow\text{x}=-26$ and $\text{y}=50$

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