# Rules of logarithm can make a big change

Calculus Level 5

$\Large \log_{x^{1/x}} a = 1 + \log_{x^{\log_m a}} a$

In the above equation, the first logarithm has base $$x^{1/x}$$, and the second logarithm has base $$x$$ raised to the power of $$\log_m a$$.

Let $$m,a$$ be real numbers where $$m > 0, a > 1$$, such that the equation above has exactly one real solution for $$x$$; suppose that this solution is $$x = n$$. Let

$\large{ P = \frac{(mn + n^2)^3 - m^3n^3 - n^6}{3 n^2 (m+n)} }$

Compute the value of $$\left\lfloor 10^3 \cdot a^P \right\rfloor$$.

×