Find integer values of $m$ and $n$ (see description)

Question

Find integer values of $m$ and $n$ for which $m-n\log_32=10\log_96$.

I tried rewriting the RHS to $\log_3$ and the LHS to $\log_9$ but it seems that in each case I always miss something. I would really appreciate if someone could write a step-by-step solution.

Answer 1

$$10\log_96=5\log_36=5(1+\log_32).$$ Thus, $$m-5=(n+5)\log_32$$ and since $\log_32\not\in\mathbb Q,$ we obtain: $m=5$ and $n=-5$.

Answer 2

$$ m = \log _32^n+ {\log_3 6^{10}\over \log _3 9}$$ so

$$ 2m = \log _32^{2n}6^{10}$$

so $$ 3^{2m}= 2^{2n}6^{10}= 2^{2n+10}3^{10}$$ so $2n+10 =0$ and $2m = 10$, so $n=-5$ and $m=5$.