Logarithms of Powers

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 $\log_a x$ be the logarithm to the base $a$ of $x$.

Then:

$\log_a \left({x^r}\right) = r \log_a x$

Proof

Natural Logarithms

Proof 1

When $a = e$, $\log_a x = \ln x$.

Consider the function $f \left({x}\right) = \ln \left({x^r}\right) - r \ln x$.

Then from:

• The definition of the natural logarithm
• The Fundamental Theorem of Calculus
• The Power Rule for Derivatives
• The Chain Rule:

$\displaystyle \forall x > 0: f^{\prime} \left({x}\right) = \frac 1 {x^r} r x^{r-1} - \frac r x = 0$

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

$\forall x > 0: \ln \left({x^r}\right) - r \ln x = c$

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \ln a\)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \iff\)	\(\displaystyle \)	\(\displaystyle e^b\)	\(=\)	\(\displaystyle a\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({e^b}\right)^c\)	\(=\)	\(\displaystyle a^c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle e^{cb}\)	\(=\)	\(\displaystyle a^c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent of Product
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \ln e^{cb}\)	\(=\)	\(\displaystyle \ln a^c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle cb\)	\(=\)	\(\displaystyle \ln a^c\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponential of Natural Logarithm

By hypothesis, $\ln a = b$.

Multiplying both sides by $c$:

$c \ln a = cb$

But we proved above that:

$cb = \ln a^c$

$\blacksquare$

General Logarithms

Let $y = r \log_a x$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a^y\)	\(=\)	\(\displaystyle a^{r \log_a x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a^{\log_a x} }\right)^r\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent Combination Laws
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^r\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of logarithm base $a$

The result follows by taking logs base $a$ of both sides.

$\blacksquare$

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.2$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 14.2 \ \text{(ii)}$
• For a video presentation of the contents of this page, visit the Khan Academy.