$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$ The problem


Given that $a,b>0$ and $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)$$
Find the value of $$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)$$

My attempt

We have from the given condition
$$2+\frac{\log a}{\log 2}=3+\frac{\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
$\implies$
$$\frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}$$
By ratio and proportion we get each ratio equal to
$$\frac{2\log 2+3\log 3+\log(ab)-\log(a+b)}{0}$$
Thus we have
$$\log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108)$$
But i am unable to find $\log(ab)$
 A: $$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$$
$$2+\log _{2} a=k \implies a=2^{k-2}$$
$$3+\log _{3} b=k \implies b=3^{k-3}$$
So, now, we have
$$\frac{\log \left(2^{k-2}+3^{k-3}\right)}{\log (6)}=k\tag 1$$ and we need to compute
$$\log _{a b}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{\log \left(2^{2-k}+3^{3-k}\right)}{\log \left(2^{k-2} 3^{k-3}\right)}\tag 2$$ There is no closed form for equation $(1)$. Using graphics and numerical methods, we find $k=-1.18196$ and plugging in $(2)$ the result is $-0.688556$.
This strange value of $k$ effectively gives
$$\log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108)$$
Strange problem ! Typo's may be.
A: Here's a way to SIMPLIFY the problem, a hint that is
We need to find $\log_{ab}{(\frac1a +\frac1b)}$
We can change this into a more workable form like this
$$\log_{ab}{(\frac1a +\frac1b)} = \log_{ab}{(\frac{a+b}{ab})} = \log_{ab}{(a+b)} - \log_{ab}{ab} = \log_{ab}{(a+b)} -1$$
So, all we have to do now is find $\log_{ab}{(a+b)} -1$ to solve your original problem. Try finding $\log_{ab}{(a+b)}$ first, then just subtract $1$. Look at your given equation now and try to solve from there.
A: Set
\begin{align}\\ \frac{2\log 2+\log a}{\log 2}=\frac{3 \log 3+\log b}{\log 3}=\frac{\log (a+b)}{\log 6}=y\\\to \log\left(\frac{1}{a}+\frac{1}{b}\right)=\log(108) \to 108ab=a+b\end{align}
This is where you are.
Let us find another equation to  eliminate  a or b.
From  \begin{align}\\2\log 2\log 3+\log a\log 3=3 \log 3\log 2+\log b\log 2\end{align} \begin{align}a=2b^{\frac{\ln 2}{\ln 3}}, 108\cdot 2b^{\frac{\ln 2+\ln 3}{\ln 3}}=2b^{\frac{\ln 2}{\ln 3}}+b\end{align}
Therefore
\begin{align}b \approx 0.01, a\approx 0.11, a+b \approx 0.12, ab\approx 0.0011
\\log_{ab}(\frac{1}{a}+\frac{1}{b})=\log_{ab}{(a+b)} -1\approx -0.6883\end{align}
