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

$$2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$$
$$2+\log _{2} a=k \implies a=2^{k-2}$$
$$3+\log _{3} b=k \implies b=3^{k-3}$$
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.
