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If $\log_{12} 27 = a$ then find the value of $\log_6 16$.

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  • $\begingroup$ This site works better if your question includes what you have tried $\endgroup$ – Henry Apr 27 '16 at 21:56
  • $\begingroup$ I didn't really get the question. $\endgroup$ – Deepansh Chellani Apr 27 '16 at 22:02
  • $\begingroup$ What do you not understand? The meaning of $\log$ with a base? $\endgroup$ – peterwhy Apr 27 '16 at 22:11
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    $\begingroup$ I am not surprised. It is a bizarre question. It seems to want $\log_616$ in terms of $a=\log_{12}27$. The answer is $\frac{4(3-a)}{3+a}$, which is far from obvious! $\endgroup$ – almagest Apr 27 '16 at 22:16
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    $\begingroup$ It's a cool question +1, particularly for a contest. (Not a beauty contest...) I don't blame the OP, it took me a while to figure it out.... $\endgroup$ – imranfat Apr 27 '16 at 22:31
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$a=\log_{12}27$ is equivalent to $3^3=12^a=2^{2a}3^a$. So $2^{2a}=3^{3-a}$. Hence $2^{3+a}=2^{3-a}3^{3-a}=6^{3-a}$. So $16=2^4=6^b$ where $b=\frac{4(3-a)}{3+a}$. Hence $\log_616=b=\frac{4(3-a)}{3+a}$.

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To solve this problem, we note that log_an⁡am=〖1+log〗_a⁡m/〖1+log〗_a⁡n and 〖(log_n⁡m)〗^(-1)=log_m⁡n

So that log_12⁡27=log_(4×3)⁡〖9×3〗=〖1+log〗_3⁡9/〖1+log〗_3⁡4 =3/(1+log_3⁡4 )=a

Hence , log_3⁡4=3/a-1.

Now, log_6⁡16=2 log_6⁡4=2/log_4⁡6 note the change in base, =2/log_4⁡〖3×2〗 =2/(log_4⁡3+log_4⁡2 )=2/(log_4⁡3+1⁄log_2⁡4 )=2/(〖(log_3⁡4)〗^(-1)+1⁄2) =2/(〖((3-a)/a)〗^(-1)+1/2)=2/(a/(3-a)+1/2)=(4(3-a))/((3+a)).

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  • $\begingroup$ I typed using MSword. $\endgroup$ – A.C.Holmes Apr 28 '16 at 8:13
  • $\begingroup$ write using tex $\endgroup$ – Nebo Alex Apr 28 '16 at 8:17

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