If $\log_{12}54=a$ then $ \log_{6}12=?$ I am given $$\log_{12}54=a$$
So what will be value of $ \log_{6}12?$
I used base changing theorem and wrote expression as 
$$\frac{\log_{6}54}{ \log_{6}12} =a$$
And then $$ \frac{1+\log_{6}9}{ a} = \log_{6}12$$ now what to do?
 A: Observe first that 
$$\begin{align}a&=\log_{12}(54)=\log_{12}(2)+3\log_{12}(3)\\&=\frac{1}{\log_2(12)}+\frac{3}{\log_3(12)}=\frac{1}{2+\log_2(3)}+\frac{3}{1+\frac{2}{\log_2(3)}}\end{align}$$
From this, we can compute $\log_2(3)$ in terms of $a$. We just need to solve for $x$ in the equation $$a=\frac{1}{2+x}+\frac{3}{1+\frac{2}{x}}$$
Then we can use this value to compute:
$$\log_6(12)=2\log_6(2)+\log_6(3)=\frac{2}{\log_2(6)}+\frac{1}{\log_3(6)}=\frac{2}{1+\log_2(3)}+\frac{1}{1+\frac{1}{\log_2(3)}}$$
The equation above is $$a=\frac{1+3x}{x+2}$$
from where $$x=\frac{2a-1}{3-a}$$
Therefore 
$$\begin{align}\log_{6}(12)&=\frac{2}{1+\frac{2a-1}{3-a}}+\frac{1}{1+\frac{1}{\frac{2a-1}{3-a}}}\\&=\frac{5}{2+a}\end{align}$$
A: $\bf hint:$
decompose the numbers $54 = 2*3^3, 12 = 2^2*3, 6 =2*3$ as use the change of basis formula $\log_a(b) = \dfrac{\ln b}{\ln a}.$ so we get 
$$a = \dfrac{\ln 2 + 3 \ln 3}{2\ln 2 + \ln 3}, b = \dfrac{2\ln 2 + \ln 3}{\ln 2 + \ln 3} $$
observe that both $a$ and $b$ depend only on the ratio $\dfrac{\ln 2}{\ln 3}.$ eliminating the ratio should give you a relation between $a$ and $b.$
