# Rules of logarithm can make a big change

**Calculus**Level 4

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

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 \(\left\lfloor 10^3 \cdot a^P \right\rfloor\).