Logarithm of Power/Natural Logarithm/Proof 3

Theorem

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:

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

Here we adopt the definition of $\ln x$ to be:

$\ds \ln x := \int_1^x \dfrac {\d t} t$

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

$=$

$\ds \int_1^{x^r} \dfrac {\d t} t$

$\ds $

$=$

$\ds \int_1^x \dfrac {r t^{r - 1} \rd t} {t^r}$

Integration by Substitution: $t \mapsto t^r$, $\d t \mapsto r t^{r - 1} \rd t$, $1 \mapsto 1$, $x^r \mapsto x$

$\ds $

$=$

$\ds r \int_1^x \dfrac {\d t} t$

$\ds $

$=$

$\ds r \ln x$

$\blacksquare$