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Question Number 165232 by Neils last updated on 27/Jan/22

Commented by bobhans last updated on 30/Jan/22

Answered by FelipeLz last updated on 27/Jan/22

$$5^{{\log }_{\frac{4}{5}}\left(5\right)}=e^{\ln \left(5\right){\log }_{\frac{4}{5}}\left(5\right)}=e^{\frac{{\ln }^2\left(5\right)}{\ln \left(4\right)-\ln \left(5\right)}}$$$$4^{{\log }_{\frac{4}{5}}\left(4\right)}=e^{\ln \left(4\right){\log }_{\frac{4}{5}}\left(4\right)}=e^{\frac{{\ln }^2\left(4\right)}{\ln \left(4\right)-\ln \left(5\right)}}$$$$\frac{e^{\frac{{\ln }^2\left(5\right)}{\ln \left(4\right)-\ln \left(5\right)}}}{e^{\frac{{\ln }^2\left(4\right)}{\ln \left(4\right)-\ln \left(5\right)}}}=e^{\frac{{\ln }^2\left(5\right)-{\ln }^2\left(4\right)}{\ln \left(4\right)-\ln \left(5\right)}}=e^{\frac{-[\ln \left(4\right)-\ln \left(5\right)][\ln \left(4\right)+\ln \left(5\right)]}{\ln \left(4\right)-\ln \left(5\right)}}=e^{-[\ln \left(4\right)+\ln \left(5\right)]}=e^{-\ln \left(20\right)}=\frac{1}{20}$$

Commented by Neils last updated on 27/Jan/22

$$NiceSolution$$

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