QUESTION on Logarithm 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.
 A: 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 $2bc=log_{48}{24^2}$
$24^2=12*48$
A: Starting from your answer of $abc = \log_{48}{12}$
$$
abc = \log_{48}{12} = \frac{ln{12}}{ln{48}}
$$
$$
abc+1 = \frac{ln{12}+ln{48}}{ln{48}} = \frac{ln(12\times48)}{ln{48}} = \frac{ln(24^2)}{ln{48}} = \frac{2ln{24}}{ln{48}} = 2\frac{ln{36}}{ln{48}}\times\frac{ln{24}}{ln{36}}
$$
$$
abc+ 1 = 2\log_{48}{36}\times\log_{36}{24} = 2bc
$$
A: 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.
