If ( a^ -1 b^2 a^2 b^ -4 )^7 ( a^3 b^ -5 a^ -2 b^3 )^ -5 =a^x b^y, find x+y

( a^ -1 b^2 a^2 b^ -4 )^7 ( a^3 b^ -5 a^ -2 b^3 )^ -5 =a^x b^y (b^6 a^3 )^7 (a^5 b^8 )^ -5 =a^x b^y (b^6 a^3 )^7 (b^8 a^5 )^5=a^x b^y ( b^ 42 a^ 21 ) ( b^ 40 a^ 25 )=a^x b^y ( b^ 42 a^ 21 ) ( a^ 25 b^ 40 )=a^x b^y b^2 a^4=a^x b^y x=4 and y=2 x+y=4+2 A=6

If ( a^ -1 b^2 a^2 b^ -4 )^7 ( a^3 b^ -5 a^ -2 b^3 )^ -5 =a^x b^y…

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Question

If $\left(\frac{a^{-1} b^2}{a^2 b^{-4}}\right)^7 \div\left(\frac{a^3 b^{-5}}{a^{-2} b^3}\right)^{-5}=a^x \cdot b^y$, find $x+y$

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Answer

$ \left(\frac{a^{-1} b^2}{a^2 b^{-4}}\right)^7 \div\left(\frac{a^3 b^{-5}}{a^{-2} b^3}\right)^{-5}=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^6}{a^3}\right)^7 \div\left(\frac{a^5}{b^8}\right)^{-5}=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^6}{a^3}\right)^7 \div\left(\frac{b^8}{a^5}\right)^5=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^{42}}{a^{21}}\right) \div\left(\frac{b^{40}}{a^{25}}\right)=a^x \cdot b^y$
$ \Rightarrow\left(\frac{b^{42}}{a^{21}}\right) \times\left(\frac{a^{25}}{b^{40}}\right)=a^x \cdot b^y$
$ \Rightarrow b^2 \times a^4=a^x \times b^y$
$ \Rightarrow x=4$ and $ y=2$
$ \Rightarrow x+y=4+2$
$A=6$