<s>[INST] How to simplify this expression using logarithms I want to find the value of this expression in as simple way as possible.
$$  \frac{1+ 2\log_3 2}{(1+ \log_3 2)^2}+ \log^2_6 2  $$
I simplified and I am stuck at 
$$ \frac{1+2\log_3 2+2\log_6 2+4\log_3 2×2\log_6 2}{1+4\log_3 2} $$
 [/INST] Knowing that $\log_ab = \frac{\log b}{\log a}$
Then the expression becomes:
$$\frac{1 + 2\frac{\log 2}{\log 3}}{(1+\frac{\log 2}{\log 3})^2} + (\frac{\log 2}{\log 6})^2$$
$$= \frac{1 + 2\frac{\log 2}{\log 3}}{1+2\frac{\log 2}{\log 3} + \frac{\log^2 2}{\log^2 3}} + (\frac{\log 2}{\log 3 + \log 2})^2$$
$$= \frac{\log^23 + 2\log 2\log 3}{\log^23+2\log2\log 3 + \log^2 2} + \frac{\log^2 2}{\log^23 + 2\log2\log3 + \log^22}$$
$$= \frac{\log^23 + 2\log 2\log 3 + \log^22}{\log^23+2\log2\log 3 + \log^2 2}$$
$$=1$$
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