Rules of logarithm can make a big change

${\text{The identity }}{\log _p}q{\text{ }} = {\text{ }}\frac{1}{{{{\log }_q}p}}{\text{ applies to the given eqation}}{\text{.}}$ ${\log _{{x^{1/x}}}}(a) = 1 + {\log _{{x^{{{\log }_m}a}}}}(a){\text{ }} \Rightarrow \frac{1}{{{{\log }_a}({x^{1/x}})}} = 1 + \frac{1}{{\mathop {\log }\nolimits_a ({x^{{{\log }_m}(a)}})}}{\text{ }} \Rightarrow \frac{1}{{\tfrac{1}{x}\mathop {\log }\nolimits_a (x)}} = 1 + \frac{1}{{\mathop {\log }\nolimits_m (a)\mathop {\log }\nolimits_a (x)}}$ $\Rightarrow x = {\log _a}(x) + {\log _a}(m)$ $\Rightarrow x = {\log _a}(mx){\text{ }} \Rightarrow {\text{ }}\boxed{{a^x} = mx}$ ${\text{This can be interpreted as the intersection between curve }}y = {a^x}{\text{ and line }}y = mx.$ ${\text{Given }}m > 0{\text{ and }}a > 1{\text{ so, one solution is only possible when}}$ ${\text{line }}y = mx{\text{ is tangent to the curve }}y = {a^x}{\text{ at }}x = n.$ ${\text{For }}y = {a^x},\frac{{dy}}{{dx}} = {a^x}\ln a$ ${\text{At }}x = n,{\text{ }}{a^n} = mn.........(1){\text{ and }}{\left( {\frac{{dy}}{{dx}}} \right)_{x = n}} = m = {a^n}\ln a.........(2)$ $(1){\text{ and (2) }} \Rightarrow {\text{ }}mn\ln a = m{\text{ }} \Rightarrow \ln a = \frac{1}{n} \Rightarrow \boxed{a = {e^{\frac{1}{n}}}}{\text{ and }}m = {\left( {{e^{\frac{1}{n}}}} \right)^n}\ln \left( {{e^{\frac{1}{n}}}} \right) \Rightarrow \boxed{m = \frac{e}{n}}$ ${\text{Now, substituting these results for }}m{\text{ and }}a{\text{ into the given expression,}}$ $\left\lfloor {{{10}^3}{a^{\frac{{{{(mn + {n^2})}^3} - {m^3}{n^3} - {n^6}}}{{3{n^2}(m + n)}}}}} \right\rfloor {\text{ }} = {\text{ }}\left\lfloor {{{10}^3}{a^{\frac{{{m^3}{n^3} + {n^6} + 3{n^4}m(m + n) - {m^3}{n^3} - {n^6}}}{{3{n^2}(m + n)}}}}} \right\rfloor {\text{ = }}\;\left\lfloor {{{10}^3}{a^{{n^2}m}}} \right\rfloor {\text{ }}\; = \left\lfloor {{{10}^3}{{\left( {{e^{\frac{1}{n}}}} \right)}^{{n^2} \times \frac{e}{n}}}} \right\rfloor {\text{ }}$ $= {\text{ }}\left\lfloor {1000{e^e}} \right\rfloor {\text{ }} = {\text{ }}\boxed{\boxed{15154}}$