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Question Number 68712 by Rio Michael last updated on 15/Sep/19

$$giventhatxandyaretwonumbersotherone.$$ $$giventhata>0andb>0$$ $$and{a}^x=b^y={\left(ab\right)}^{xy}showthatx+y=0$$

Commented byPrithwish sen last updated on 15/Sep/19

$$\mathrm{Let}{\mathrm{a}}^{\mathrm{x}}={\mathrm{b}}^{\mathrm{y}}={\left(\mathrm{ab}\right)}^{\mathrm{xy}}=\mathbf{k}$$ $$\therefore \mathbf{a}={\mathbf{k}}^{\frac{1}{\mathbf{x}}}$$ $$\mathbf{b}={\mathbf{k}}^{\frac{1}{\mathbf{y}}}$$ $$\mathbf{ab}={\mathbf{k}}^{\frac{1}{\mathbf{xy}}}$$ $$\therefore {\mathrm{k}}^{\frac{1}{\mathrm{x}}}.{\mathrm{k}}^{\frac{1}{\mathrm{y}}}={\mathrm{k}}^{\frac{1}{\mathrm{xy}}}\Rightarrow \frac{1}{\mathbf{x}}+\frac{1}{\mathbf{y}}=\frac{1}{\mathbf{xy}}\Rightarrow \mathbf{x}+\mathbf{y}=1$$ $$\mathbf{please}\mathbf{check}$$

Commented byMJS last updated on 15/Sep/19

$$a^0+b^0={\left(ab\right)}^0\Rightarrow x=y=0$$ $$\mathrm{is}\mathrm{one}\mathrm{solution}$$ $$a^x=b^y\Rightarrow y=\frac{\ln a}{\ln b}x$$ $$a^x={\left(ab\right)}^{xy}\Rightarrow y=\frac{\ln a}{\ln a+\ln b}$$ $$\frac{\ln a}{\ln b}x=\frac{\ln a}{\ln a+\ln b}\Rightarrow x=\frac{\ln b}{\ln a+\ln b}$$ $$\Rightarrow x+y=1$$

Commented byPrithwish sen last updated on 15/Sep/19

$$\mathrm{Thanks}\mathrm{Sir}.$$

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