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Question Number 143812 by liberty last updated on 18/Jun/21

$$\log {}_{\mathrm{a}}\left(\mathrm{ax}\right).\log {}_{\mathrm{x}}\left(\mathrm{ax}\right)=\log {}_{{\mathrm{a}}^2}\left(\frac{1}{\mathrm{a}}\right)$$ $$\mathrm{a}>0,\mathrm{a}\ne 1.\mathrm{So}\mathrm{x}=?$$

Answered by Olaf_Thorendsen last updated on 18/Jun/21

$${\log }_a\left(ax\right).{\log }_x\left(ax\right)={\log }_{a^2}\left(\frac{1}{a}\right)$$ $$\frac{\ln \left(ax\right)}{\ln a}.\frac{\ln \left(ax\right)}{\ln x}=-\frac{1}{2}{\log }_{a^2}{a}^2=-\frac{1}{2}$$ $${\ln }^2\left(ax\right)=-\frac{1}{2}\ln a.\ln x$$ $${\ln }^{2}a+{\ln }^{2}x+2\ln a.\ln x=-\frac{1}{2}\ln a.\ln x$$ $${\ln }^{2}x+{\ln }^{2}a+\frac{5}{2}\ln a.\ln x=0$$ $${(\ln x+\frac{5}{4}\ln a)}^2=\frac{9}{16}{\ln }^{2}a$$ $$\ln x+\frac{5}{4}\ln a=\pm \frac{3}{4}\mid \ln a\mid =\pm \frac{3}{4}\ln a$$ $$\ln x=-\frac{5}{4}\ln a\pm \frac{3}{4}\ln a$$ $$\ln x=-2\ln a\mathrm{or}-\frac{1}{2}\ln a$$ $$x=\frac{1}{a^2}\mathrm{or}\frac{1}{\sqrt{a}}$$

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