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Mathematics

If (a−1b2a2b−4)7÷(a3b−5a−2b3)−5=ax.by\Big(\dfrac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\dfrac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y, find x + y.

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Answer

Given,

⇒(a−1b2a2b−4)7÷(a3b−5a−2b3)−5=ax.by⇒(a−1−2b2−(−4))7÷(a3−(−2)b−5−3)−5=ax.by⇒(a−3b6)7÷(a5b−8)−5=ax.by⇒(a−3×7.b6×7)÷(a5×−5.b−8×−5)=ax.by⇒(a−21.b42)÷(a−25.b40)=ax.by⇒a−21.b42a−25.b40=axby⇒a−21−(−25).b42−40=axby⇒a−21+25.b2=ax.by⇒a4.b2=ax.by⇒x=4 and y=2.\Rightarrow \Big(\dfrac{a^{-1}b^2}{a^2b^{-4}}\Big)^7 ÷ \Big(\dfrac{a^3b^{-5}}{a^{-2}b^3}\Big)^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-1 - 2}b^{2 - (-4)})^7 ÷ (a^{3 - (-2)}b^{-5 - 3})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3}b^6)^7 ÷ (a^5b^{-8})^{-5} = a^x.b^y \\[1em] \Rightarrow (a^{-3 \times 7}.b^{6 \times 7}) ÷ (a^{5 \times -5}.b^{-8 \times -5}) = a^x.b^y \\[1em] \Rightarrow (a^{-21}.b^{42}) ÷ (a^{-25}.b^{40}) = a^x.b^y \\[1em] \Rightarrow \dfrac{a^{-21}.b^{42}}{a^{-25}.b^{40}} = a^xb^y \\[1em] \Rightarrow a^{-21 - (-25)}.b^{42 - 40} = a^xb^y \\[1em] \Rightarrow a^{-21 + 25}.b^{2} = a^x.b^y \\[1em] \Rightarrow a^4.b^2 = a^x.b^y \\[1em] \Rightarrow x = 4 \text{ and } y = 2.

x + y = 4 + 2 = 6.

Hence, x + y = 6.

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