Equation in Logarithm

Question

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.

Answer 1

You can use the change of base formula: $\log_a(b)=\frac{\ln(b)}{\ln(a)}$.

Answer 2

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}$

Answer 3

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

$\large c=\frac{1}{2ab-1}$