Logarithm of Power/Natural Logarithm/Proof 1
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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
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:
- $c = \ln 1 - r \ln 1 = 0$
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.2 \ \text{(ii)}$
- For a video presentation of the contents of this page, visit the Khan Academy.