If $\log_23 = a$ and $\log_52=b$ then $\log_{24}50$ is equal to?

Question

I guess this has to be done by using simple logarithmic rules, but I do not how to start. Answer in my booklet is ${b+2}\over{b(a+3)}$

Answer 1

hint: $\log_{24}50 = \log_{24}(2\cdot 5^2) = \log_{24}2+2\log_{24}5=\dfrac{1}{\log_{2}24}+\dfrac{2}{\log_{5}24}=\dfrac{1}{3+\log_{2}3}+\dfrac{2}{3\log_{5}2+\log_{5}2\cdot \log_{2}3}=\dfrac{1}{a+3}+\dfrac{2}{3b+ab}=...$

Answer 2

Clearly, $3=2^a, 5=2^{1/b}$, which shows

$24=2^3\cdot 3=2^{a+3},~50=2\cdot 5^2=2^{1+2/b}=2^{(b+2)/b}$

So $50=24^{(b+2)/(b(a+3))}$

QED