If $\log_{12} 27 = a$ then find the value of $\log_6 16$.
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This site works better if your question includes what you have tried – Henry Apr 27 '16 at 21:56
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I didn't really get the question. – Deepansh Chellani Apr 27 '16 at 22:02
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What do you not understand? The meaning of $\log$ with a base? – peterwhy Apr 27 '16 at 22:11
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5I 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! – almagest Apr 27 '16 at 22:16
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1It'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.... – 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 $\log_616=b=\frac{4(3-a)}{3+a}$.
almagest
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To solve this problem, we note that log_anam=〖1+log〗_am/〖1+log〗_an and 〖(log_nm)〗^(-1)=log_mn
So that log_1227=log_(4×3)〖9×3〗=〖1+log〗_39/〖1+log〗_34 =3/(1+log_34 )=a
Hence , log_34=3/a-1.
Now, log_616=2 log_64=2/log_46 note the change in base, =2/log_4〖3×2〗 =2/(log_43+log_42 )=2/(log_43+1⁄log_24 )=2/(〖(log_34)〗^(-1)+1⁄2) =2/(〖((3-a)/a)〗^(-1)+1/2)=2/(a/(3-a)+1/2)=(4(3-a))/((3+a)).
