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Question Number 197419 by universe last updated on 16/Sep/23

Answered by cortano12 last updated on 17/Sep/23

$$\{\begin{array}{l}{\left(2\mathrm{a}\right)}^{\ln \mathrm{a}}={\left(\mathrm{bc}\right)}^{\ln \mathrm{b}}\\ {\mathrm{b}}^{\ln 2}={\mathrm{a}}^{\ln \mathrm{c}}\end{array}$$$$\{\begin{array}{l}\ln \left(2\mathrm{a}\right).\ln \left(\mathrm{a}\right)=\ln \left(\mathrm{b}\right).\ln \left(\mathrm{bc}\right)\\ \ln \left(\mathrm{b}\right).\ln \left(2\right)=\ln \left(\mathrm{c}\right).\ln \left(\mathrm{a}\right)\end{array}$$$$\Rightarrow \ln \mathrm{a}\{\ln 2+\ln \mathrm{a}\}=\ln \mathrm{b}\{\ln \mathrm{b}+\ln \mathrm{c}\}$$$$\ln \mathrm{a}\{\frac{\ln \mathrm{c}.\ln \mathrm{a}}{\ln \mathrm{b}}+\ln \mathrm{a}\}=\ln \mathrm{b}\{\ln \mathrm{b}+\ln \mathrm{c}\}$$$$\ln {}^2\mathrm{a}\{\ln \mathrm{c}+\ln \mathrm{b}\}=\ln {}^2\mathrm{b}\{\ln \mathrm{b}+\ln \mathrm{c}\}$$$$(\ln \mathrm{c}+\ln \mathrm{b})(\ln \mathrm{a}-\ln \mathrm{b})(\ln \mathrm{a}+\ln \mathrm{b})=0$$$$\{\begin{array}{l}\ln \mathrm{c}=-\ln \mathrm{b}\Rightarrow \mathrm{bc}=1\\ \ln \mathrm{a}=0\Rightarrow \mathrm{a}=1\end{array}$$$$$$

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