Logarithm of Power/Natural Logarithm/Proof 1

Theorem

Let $x \in \R$ be a strictly positive real number.

Let $a \in \R$ be a real number such that $a > 1$.

Let $r \in \R$ be any real number.

Let $\ln x$ be the natural logarithm of $x$.

Then:

$\map \ln {x^r} = r \ln x$

Proof

Consider the function $\map f x = \map \ln {x^r} - r \ln x$.

Then from:

The definition of the natural logarithm
The Fundamental Theorem of Calculus
The Power Rule for Derivatives
The Chain Rule for Derivatives:
$\forall x > 0: \map {f'} x = \dfrac 1 {x^r} r x^{r-1} - \dfrac r x = 0$

Thus from Zero Derivative implies Constant Function, $f$ is constant:

$\forall x > 0: \map \ln {x^r} - r \ln x = c$

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

From Logarithm of 1 is 0:

$\ln 1 = 0$

Thus:

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

$\blacksquare$