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Question Number 196690 by MATHEMATICSAM last updated on 29/Aug/23

$$\mathrm{If}{\left(ax\right)}^{\log a}={\left(bx\right)}^{\log b}\mathrm{then}\mathrm{prove}\mathrm{that}$$$$x=\frac{1}{ab}.$$

Answered by BaliramKumar last updated on 29/Aug/23

$$\mathrm{loga}\cdot \log \left(\mathrm{ax}\right)=\mathrm{logb}\cdot \log \left(\mathrm{bx}\right)$$$$\mathrm{loga}\cdot (\mathrm{loga}+\mathrm{logx})=\mathrm{logb}\cdot (\mathrm{logb}+\mathrm{logx})$$$${\log }^2\mathrm{a}+\mathrm{loga}\cdot \mathrm{logx}={\log }^2\mathrm{b}+\mathrm{logb}\cdot \mathrm{logx}$$$$\mathrm{loga}\cdot \mathrm{logx}-\mathrm{logb}\cdot \mathrm{logx}={\log }^2\mathrm{b}-{\log }^2\mathrm{a}$$$$\mathrm{logx}(\mathrm{loga}-\mathrm{logb})=(\mathrm{logb}-\mathrm{loga})(\mathrm{logb}+\mathrm{loga})$$$$\mathrm{logx}\cdot \cancel{\log \left(\frac{\mathrm{a}}{\mathrm{b}}\right)}=-\cancel{\log \left(\frac{\mathrm{a}}{\mathrm{b}}\right)}\cdot \mathrm{logab}$$$$\mathrm{logx}=-\log \left(\mathrm{ab}\right)$$$$\mathrm{logx}=\log \left(\frac{1}{\mathrm{ab}}\right)$$$$\mathrm{x}=\frac{1}{\mathrm{ab}}$$

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