Logarithm of Power/Natural Logarithm

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 1

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:

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

$\blacksquare$

Proof 2

\(\ds \ln a\)

\(=\)

\(\ds b\)

\(\ds \leadstoandfrom \ \ \)

\(\ds e^b\)

\(=\)

\(\ds a\)

\(\ds \leadsto \ \ \)

\(\ds \paren {e^b}^c\)

\(=\)

\(\ds a^c\)

\(\ds \leadsto \ \ \)

\(\ds e^{c b}\)

\(=\)

\(\ds a^c\)

Exponential of Product

\(\ds \leadsto \ \ \)

\(\ds \ln e^{c b}\)

\(=\)

\(\ds \ln a^c\)

\(\ds \leadsto \ \ \)

\(\ds c b\)

\(=\)

\(\ds \ln a^c\)

Exponential of Natural Logarithm

By hypothesis, $\ln a = b$.

Multiplying both sides by $c$:

$c \ln a = c b$

But we proved above that:

$c b = \ln a^c$

$\blacksquare$

Proof 3

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

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

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

\(=\)

\(\ds \int_1^{x^r} \dfrac {\d t} t\)

Definition of Natural Logarithm

\(\ds \)

\(=\)

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

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

\(\ds \)

\(=\)

\(\ds r \int_1^x \dfrac {\d t} t\)

Primitive of Constant Multiple of Function

\(\ds \)

\(=\)

\(\ds r \ln x\)

Definition of Natural Logarithm

$\blacksquare$

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.