Logarithm of Power/Natural Logarithm/Proof 2
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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
\(\ds \ln a\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds e^b\) | \(=\) | \(\ds a\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {e^b}^c\) | \(=\) | \(\ds a^c\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{c b}\) | \(=\) | \(\ds a^c\) | 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:
- $c b = \ln a^c$
$\blacksquare$
Sources
- For a video presentation of the contents of this page, visit the Khan Academy.