Logarithm of Power

= Natural Logarithms =

Theorem
Let $$x \in \mathbb{R}$$ be a strictly positive real number.

Let $$r \in \mathbb{Q}$$ be a rational number.

Then $$\ln \left({x^r}\right) = r \ln x$$.

Proof
Consider the function $$f \left({x}\right) = \ln \left({x^r}\right) - r \ln x$$.

Then from:
 * The definition of the natural logarithm,
 * The Fundamental Theorem of Calculus;
 * The Power Rule for Derivatives, and
 * The Chain Rule:

$$\forall x > 0: f^{\prime} \left({x}\right) = \frac 1 {x^r} r x^{r-1} - \frac r x = 0$$.

Thus from Zero Derivative means Constant Function, $$f$$ is constant: $$\forall x > 0: \ln \left({x^r}\right) - r \ln x = c$$.

To determine the value of $$c$$, put $$x = 1$$.

From Basic Properties of Natural Logarithm, $$\ln 1 = 0$$.

Thus $$c = \ln 1 - r \ln 1 = 0$$.