# How to express log52\log_5 2 in terms of a and b (Refer to qn)

In my textbook, I came across this interesting question which I am currently struggling to solve:

If $\log_6 2 = a$ and $\log_5 3 = b$, express $\log_5 2$ in terms of a and b

The solution given is $\frac{ab}{1-a}$ but I do not know the working behind this. What is it?

## Answer

An alternative, although I agree with the exponential approach, too!

We’re given: $\log_6 2 = a$ and $\log_5 3 = b$

We want: $\log_5 2$

We must recall our logarithms rules. There are too many bases happening here, so let’s fix that!

The change of base formula gives us $\log_6 2 = \frac{\log_5 2}{\log_5 6}$ — I thought to do this because we’re looking for all base 5. Interesting! It has what we’re looking for, i.e. $\log_5 2$ !

Another rule from logarithms tells us: $\log_5 6=\log_5(2*3)=\log_5 2 + \log_5 3$. Aha! We see again our lovely longed for $\log_5 2$!!

Reviewing what we have: $a=\log_6 2 = \frac{\log_5 2}{\log_5 6}$ and $\log_5 6=\log_5 2 + \log_5 3$ $(=\log_5 2 + b)$

So, we have from the first part $a\log_5 6=\log_5 2$ — we have a substitution from above we can do!

$a\log_5 6=\log_5 2$ => $a(\log_5 2 + b)=\log_5 2$. Distributing appeals to you. And factoring somewhere down the road.

Do you see how to arrive at the final solution?

Attribution
Source : Link , Question Author : MetaKnight35 , Answer Author : Zach Haney