Equation in Logarithm We have 
${\log_45} = a $
${\log_56}= b $
and we have to find ${\log_32} $.
I tried the question but because of the different bases I was not able to get the solution.
 A: Please verify my answer.
I used : $\log_a(b)=\frac{\log(a)}{\log(b)}$ this formula and got 
${\log_56} = b $ 
$= {\log_53} + {\log_52} $ = $\frac{\log(3)}{\log(5)} +  \frac{\log(2)}{\log(5)}$= b
$ \log3 + \log 2 = b * \log5 $          this is our equation 1
Then we have 
$\log_45 = a$ 
= $\frac{\log(5)}{\log(4)} = a$
= $\frac{\log(5)}{2*\log(2)}= a$
= $\log 5 = 2* \log2 *a$  this end with second equationa after solving the equation i got
$\frac{\log(3)}{\log(2} = 2ab - 1$ 
=$ \log_32 = \frac{1}{2ab - 1}$
A: I will use a different approach, and will end up with the same solution that you have reached, thus verifying you answer.
We are given $\log_45=a$ and $\log_56=b$, and wish to find $\log_32$. 
So, $4^a=5$, $5^b=6$, $3^c=2$, and the objective is to find $c$ in terms of $a$ and $b$.
Thus $5^b=(4^a)^b=4^{ab}=6=3\cdot2$
$4^{ab}=2^{2ab}=3\cdot2$
leading to
$3=2^{2ab-1}$
Raising both sides to the power $\frac{1}{(2ab-1)}$, we end up with
$\large 3^{\frac{1}{2ab-1}}=2$
Resulting in
$\large c=\frac{1}{2ab-1}$
A: You can use the change of base formula: $\log_a(b)=\frac{\ln(b)}{\ln(a)}$.
