Logarithm of Power/Natural Logarithm/Natural Power

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

Let $n \in \R$ be any natural number.

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

Then:
 * $\map \ln {x^n} = n \ln x$

Proof
Proof by Mathematical Induction:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\forall x \in \R_{>0}: \map \ln {x^n} = n \ln x$

Basis for the Induction
$\map P 0$ is the case:
 * $\forall x \in \R_{>0}: \map \ln 1 = 0$

from Logarithm of 1 is 0.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\forall x \in \R_{>0}: \map \ln {x^k} = k \ln x$

Then we need to show:
 * $\forall x \in \R_{>0}: \map \ln {x^{k + 1} } = \paren {k + 1} \ln x$

Inductive Step
This is our induction step:

Fix $x \in \R_{>0}$.

Then:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: \map \ln {x^n} = n \ln x$