Find $x$ if $$\log_ax=5\log_am+\dfrac12\left[\log_a(m+n)+\dfrac13\log_a(m-n)-\log_am-\log_an\right]$$
The idea is to write the RHS as $\log_a T$, right? Then we will have $\log_ax=\log_aT\Rightarrow x=T$.
The RHS is $$5\log_am+\dfrac12\log_a(m+n)+\dfrac16\log_a(m-n)-\dfrac12\log_a(mn)=\\\log_am^5+\log_a(m+n)^\frac12+\log_a(m-n)^\frac16-\log_a(mn)^\frac12\\=\log_a\dfrac{m^5(m+n)^\frac12(m-n)^\frac16}{(mn)^\frac12}=\log_a\dfrac{m^\frac92(m+n)^\frac12(m-n)^\frac16}{n^\frac12}$$ Can we write the last expression in more simplified form? Would you leave it like this?
I simplified further to $$x=\dfrac{\sqrt{\left(m^4\right)^2.m}\sqrt{m+n}\sqrt[6]{m-n}}{\sqrt{n}}=\dfrac{m^4\sqrt{m(m+n)}\sqrt[6]{m-n}}{\sqrt{n}}\\=\dfrac{m^4\sqrt{mn(m+n)}\sqrt[6]{m-n}}{n}=\dfrac{m^4\sqrt[6]{m^3n^3(m+n)^3(m-n)}}{n}$$
