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

\log_{a} (ax) . \log_{x} (ax) = \log_{a^{2}} (\frac{1}{a})

a > 0, a \neq 1 . So x = ?

Answered by Olaf_Thorendsen last updated on 18/Jun/21

\log_{a} (a x) . \log_{x} (a x) = \log_{a^{2}} (\frac{1}{a})

\frac{\ln (a x)}{\ln a} . \frac{\ln (a x)}{\ln x} = - \frac{1}{2} \log_{a^{2}} a^{2} = - \frac{1}{2}

\ln^{2} (a x) = - \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} ∣ \ln a ∣ = \pm \frac{3}{4} \ln a

\ln x = - \frac{5}{4} \ln a \pm \frac{3}{4} \ln a

\ln x = - 2 \ln a or - \frac{1}{2} \ln a

x = \frac{1}{a^{2}} or \frac{1}{\sqrt{a}}