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.

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.
