Logarithm of Power/Natural Logarithm/Natural Power
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Theorem
Let $x \in \R$ be a strictly positive real number.
Let $n \in \Z_{\ge 0}$ be any natural number.
Let $\ln x$ be the natural logarithm of $x$.
Then:
- $\map \ln {x^n} = n \ln x$
Proof
Proof by Mathematical Induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\forall x \in \R_{>0}: \map \ln {x^n} = n \ln x$
Basis for the Induction
$\map P 0$ is the case:
- $\forall x \in \R_{>0}: \map \ln 1 = 0$
from Logarithm of 1 is 0.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\forall x \in \R_{>0}: \map \ln {x^k} = k \ln x$
Then we need to show:
- $\forall x \in \R_{>0}: \map \ln {x^{k + 1} } = \paren {k + 1} \ln x$
Induction Step
This is our induction step:
Fix $x \in \R_{>0}$.
Then:
\(\ds \map \ln {x^{k + 1} }\) | \(=\) | \(\ds \map \ln {x^k x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \ln {x^k} + \ln x\) | Sum of Logarithms/Natural Logarithm | |||||||||||
\(\ds \) | \(=\) | \(\ds k \ln x + \ln x\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1} \ln x\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: \map \ln {x^n} = n \ln x$
$\blacksquare$