If a6b−43=ax.b2y\sqrt[3]{a^6b^{-4}} = a^x.b^{2y}3a6b−4=ax.b2y, find x and y, where a, b are different positive primes.
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Given,
⇒a6b−43=ax.b2y⇒(a6b−4)13=ax.b2y⇒a63.b−43=ax.b2y⇒a2.b−43=ax.b2y∴x=2 and 2y=−43⇒x=2 and y=−23.\Rightarrow \sqrt[3]{a^6b^{-4}} = a^x.b^{2y} \\[1em] \Rightarrow (a^6b^{-4})^{\dfrac{1}{3}} = a^x.b^{2y} \\[1em] \Rightarrow a^{\dfrac{6}{3}}.b^{-\dfrac{4}{3}} = a^x.b^{2y} \\[1em] \Rightarrow a^2.b^{-\dfrac{4}{3}} = a^x.b^{2y} \\[1em] \therefore x = 2 \text{ and } 2y = -\dfrac{4}{3} \\[1em] \Rightarrow x = 2 \text{ and } y = -\dfrac{2}{3}.⇒3a6b−4=ax.b2y⇒(a6b−4)31=ax.b2y⇒a36.b−34=ax.b2y⇒a2.b−34=ax.b2y∴x=2 and 2y=−34⇒x=2 and y=−32.
Hence, x = 2 and y = −23-\dfrac{2}{3}−32.
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