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How do I solve the following equation:

$(3x)^{ln3}=(4x)^{ln4}$

Thanks in advance!

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    $\begingroup$ Take the logarithms of both sides. $\endgroup$ Feb 4, 2016 at 14:15
  • $\begingroup$ @Sheow Boon Do you know what is $a^{ln a}$ ? $\endgroup$
    – Nikunj
    Feb 4, 2016 at 14:19

3 Answers 3

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Take the natural logarithms of both sides, then \begin{equation} \ln 3 (\ln 3 + \ln x)=\ln 4 ( \ln 4 + \ln x) \end{equation} Thus \begin{equation} \ln x =\frac{(\ln 3)^2 - (\ln 4)^2}{\ln 4 - \ln 3}=-(\ln 4+\ln 3)=-\ln 12=\ln \frac{1}{12}. \end{equation} Since $\ln x$ is injective, $x=\frac{1}{12}$.

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$$(3x)^{\ln(3)}=(4x)^{\ln(4)} \quad \iff\quad \frac{3^{\ln(3)}}{4^{\ln(4)}}=x^{\ln(4)-\ln(3)}\quad \iff\quad x = \bigg(\frac{3^{\ln(3)}}{4^{\ln(4)}}\bigg)^{\frac1{\ln(4)-\ln(3)}}$$

If you want a "nicer" solution. As pointed out in the comments, take the logarithm: We have $$\ln(x)=\ln\Bigg(\bigg(\frac{3^{\ln(3)}}{4^{\ln(4)}}\bigg)^{\frac1{\ln(4)-\ln(3)}}\Bigg)=\frac{\ln(3)^2-\ln(4)^2}{\ln(4)-\ln(3)}\\=\frac{\big(\ln(3)-\ln(4)\big)\big(\ln(3)+\ln(4)\big)}{\ln(4)-\ln(3)}=-(\ln(4)+\ln(3))=\ln\bigg(\frac{1}{12}\bigg)$$ And thus, $x=1/12$ by taking the exponential on both sides.

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This doesn't require using logarithms to get an answer.

$$(3x)^{\ln 3}=(4x)^{\ln 4}$$ $$3^{\ln 3}x^{\ln 3} = 4^{\ln 4}x^{\ln 4}$$ $$\frac{x^{\ln 4}}{x^{\ln 3}} = \frac{3^{\ln 3}}{4^{\ln 4}}$$ $$x^{\ln 4 - \ln 3} = \frac{3^{\ln 3}}{4^{\ln 4}}$$ $$x^{\ln \frac43} = \frac{3^{\ln 3}}{4^{\ln 4}}$$ $$x = \left(\frac{3^{\ln 3}}{4^{\ln 4}}\right)^{\frac{1}{\ln \frac43}}$$

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  • $\begingroup$ See other answers for details of the simplification. I don't want to just duplicate their answers here. $\endgroup$
    – MPW
    Feb 4, 2016 at 15:45

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