1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 5
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

$a=\log_{12}27$ 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 $\log_616=b=\frac{4(3-a)}{3+a}$.

$\endgroup$
0
$\begingroup$

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)).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.