Logarithm of Power

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

Proof 2
By hypothesis, $\ln a = b$.

Multiplying both sides by $c$:


 * $c \ln a = cb$

But we proved above that:


 * $cb = \ln a^c$

General Logarithms
Let $y = r \log_a x$.

Then:

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