2
$\begingroup$

This question was easy to solve, but when it came to matching options I failed. Hope you can help. Let,

  • $a= \log_{24}(12)$
  • $b= \log_{48}(36)$
  • $c= \log_{36}(24)$

then $abc +1 = ?$

  • a) $2ab$
  • b) $2bc$
  • c) $2ca$
  • d) $ba + bc$

FYI :- I tried to solve and found $abc= log_{48}(12)$ and did several things but none were useful to get the right option I tried to take $24^a$ like that, but it didn't help me. Hope you can help. The answer was (b) but I didn't get it.

$\endgroup$
  • $\begingroup$ So you mean $a = \log_{24}(12)$ ? $\endgroup$ – Zubzub Apr 13 '17 at 15:11
  • $\begingroup$ Yes that only @Zubzub $\endgroup$ – TreaV Apr 13 '17 at 15:12
  • 1
    $\begingroup$ Use proofwiki.org/wiki/Change_of_Base_of_Logarithm $\endgroup$ – lab bhattacharjee Apr 13 '17 at 15:15
  • $\begingroup$ Hopefully my edit is right ! $\endgroup$ – Zubzub Apr 13 '17 at 15:16
  • $\begingroup$ @Zubzub thanks for edit , $\endgroup$ – TreaV Apr 13 '17 at 15:19
3
$\begingroup$

It is the second option.(b) As you said, $abc = log_{48}{12}$

And we know that $log_{a}b+1=log_{a}b+log_{a}a=log_{a}ab$ so $abc + 1 =$

$log_{48}{12 * 48}$

On the other hand the second option is: $2bc = 2*log_{48}{}{24}$ and we know that $a*log_{b}{c} = log_{b}{c^a}$ so $2bc=log_{48}{24^2}$

$24^2=12*48$

$\endgroup$
1
$\begingroup$

Starting from your answer of $abc = \log_{48}{12}$ $$ abc = \log_{48}{12} = \frac{ln{12}}{ln{48}} $$ $$ abc+1 = \frac{ln{12}+ln{48}}{ln{48}} = \frac{ln(12\times48)}{ln{48}} = \frac{ln(24^2)}{ln{48}} = \frac{2ln{24}}{ln{48}} = 2\frac{ln{36}}{ln{48}}\times\frac{ln{24}}{ln{36}} $$ $$ abc+ 1 = 2\log_{48}{36}\times\log_{36}{24} = 2bc $$

$\endgroup$
0
$\begingroup$

It looks like you already know that $\log_a(b) = \frac{\log a}{\log b}$.

Since all of the numbers have $2$ and $3$ as their only prime factors, we can start by setting $x=\log 2$ and $y=\log 3$. We then have

$$ a = \log_{24}{12} = \frac{2x+y}{3x+y} \qquad b = \log_{48}{36} = \frac{2x+2y}{4x+y} \qquad c = \log_{36}{24} = \frac{3x+y}{2x+2y} $$

You can then compute $$abc+1 = \frac{2x+y}{4x+y} + 1 = \frac{6x+2y}{4x+y} $$

Do simiarly for each of the the answer options and see if one of them matches. For option (b) we get $$ 2bc = 2\frac{2x+2y}{4x+y}\cdot\frac{3x+y}{2x+2y} = 2\frac{3x+y}{4x+y}=\frac{6x+2y}{4x+y} $$ which is the result we're searching for.

$\endgroup$

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.