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Given :

$$\log_{12}18 = a \text{ and }\log_{24}54=b$$

prove that:

$$ab + 5(a-b) = 1$$

My attempt: I couldn't solve it in any way, as base were not common. I could solve it if base of second equation was $12$ raised to something, but here it is $12\times 2$ :( Any help will be greatly appreciated.

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  • $\begingroup$ ..."prove this logarithm"?? Huh? $\endgroup$ – JP McCarthy May 13 '15 at 16:49
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HINT:

$$a=\log_{12}18=\dfrac{\log(18)}{\log(12)}=\dfrac{\log2+2\log3}{2\log2+\log3}$$

as $\log_ab=\dfrac{\log b}{\log a}$ and $\log(a^2b)=\log(a^2)+\log(b)=2\log a+\log b$

Express $\log2$ in terms of $\log3$

Similarly for $b$

Then compare the values of $\log2$ to eliminate it

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Note that $a=\frac{log(18)}{log(12)}$ and $b=\frac{log(54)}{log(24)}$

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  • $\begingroup$ Oh it was so simple! many thanks! $\endgroup$ – Max Payne May 13 '15 at 16:52

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