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If $\log_{12}18=a$, then what is $\log_{24}16$?

I was solving this and couldn't reach the result.

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closed as off-topic by Vidyanshu Mishra, user91500, Claude Leibovici, M. Winter, mlc Jun 6 '17 at 15:36

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Alternative hint:  using $\;\log_a b = \cfrac{\ln b}{\ln a}\;$ it follows that:

  • $a = \log_{12}{18}=\cfrac{\ln 18}{\ln 12}= \cfrac{\ln 2 \cdot 3^2}{\ln 2^2\cdot 3} = \cfrac{\ln 2 + 2 \ln 3}{2 \ln 2 + \ln 3}=\cfrac{1 + 2 \cfrac{\ln 3}{\ln 2}}{2 + \cfrac{\ln 3}{\ln 2}}=\cfrac{1 + 2 \log_2 3}{2 + \log_2 3}$

  • $\log_{24}{16}=\cfrac{\ln 16}{\ln 24}=\cfrac{\ln 2^4}{\ln 2^3 \cdot 3}=\cfrac{4 \ln 2}{3 \ln 2 + \ln 3}=\cfrac{4}{3+\cfrac{\ln 3}{\ln 2}}=\cfrac{4}{3 + \log_2 3}$

Eliminating $\,\log_2 3\,$ between the two expressions gives $\,\log_{24}{16}\,$ in terms of $\,a\,$.

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    $\begingroup$ Thank you so much for your help. $\endgroup$ – Iti Shree May 14 '17 at 1:32
  • $\begingroup$ @user135711 Thanks for catching that. Edited and fixed. $\endgroup$ – dxiv May 21 '17 at 16:04
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Hint:

Use the two identities:

$$\begin{align}\log_{b} c &=\frac{1}{\log_c b}\\ \log_b(cd)&=\log_b c +\log_b d\end{align}$$

To get $a$ in terms of $\log_2 3$ and $\log_{24} 16$ in terms of $\log_2 3$.

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  • $\begingroup$ Thank you so much it took me time but with the hint I did it ^_^ $\endgroup$ – Iti Shree May 14 '17 at 1:21

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