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?
On
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?
I find exponentials easier to deal with than logs.
The first log equation says that $2=6^a$, and the second says that $3=5^b$.
Rewrite $2=6^a$ as $2=2^a\cdot 3^a$, and then as $2=2^a(5^b)^a$. We obtain the equation $$2=2^a\cdot 5^{ab}.$$ This can be rewritten as $$2^{1-a}=5^{ab}.$$ Taking logs to the base $5$,we get $(1-a)\log_5 2=ab$. Since $a\ne 1$, the result follows.