Log3 / log2 = a and log6 / log5 = b, find log9 / log10 in terms of a and b

I've tried almost everything :(

Any help would be appreciated

Firstly we need to reevaluate some of the log terms here in order to simplify this question.
log 6 = log 3 + log 2.
log 9 = 2 log 3
log 10 = log 5 + log 2.
Thus log 9 / log 10 = 2 log 3 / (log 5 + log 2).

Rewriting the given log relationships:
log 3 = a log 2.
log 5 = (1/b) log 6 = (1/b)(log 3 + log 2) = (1/b)(log 2)(a+1).

We can then substitute these into the expression for log 9 / log 10:
2 log 3 / (log 5 + log 2) = 2a log 2 / [(1/b)(log 2)(a+1) + log 2] = 2a log 2 / [(1 + (1/b)(a+1))(log 2)]

Cancelling log 2 from the numerator and denominator:
log 9 / log 10 = 2a / [1 + (1/b)(a+1)].

First, you need to factor 9 into 3x3 or 3² and 10 into 2x5

So you need log 3, log 2 and log 5 from the given formulae.

Multiply the first equation throughout by log 2:
log 3 = a log 2
Divide this equation throughout by a:
log 2 = (log 3) / a

Use similar techniques to get
log5 = (log 6)/b

Now for the last one:
log 9 / log 10
= log (3²) / log (5*2)
= 2log3 / [ log 5 + log 2 ]
= 2 (a log 2) / [ (log 6 / b) + (log 3 / a) ]
= 2a log 2 / [ (log 6 / b) + (log 3 / a) ]