Logarithm of Power/Natural Logarithm/Natural Power

Theorem

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

Let $n \in \Z_{\ge 0}$ 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:

$\forall x \in \R_{>0}: \map \ln {x^{k + 1} } = \paren {k + 1} \ln x$

Induction Step

This is our induction step:

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

Then:

\(\ds \map \ln {x^{k + 1} }\)

\(=\)

\(\ds \map \ln {x^k x}\)

\(\ds \)

\(=\)

\(\ds \map \ln {x^k} + \ln x\)

Sum of Logarithms/Natural Logarithm

\(\ds \)

\(=\)

\(\ds k \ln x + \ln x\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds \paren {k + 1} \ln x\)

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

Therefore:

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

$\blacksquare$