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Log Of Powers

Theorem: \:\:log{(x^y)} = y\; log{(x)}

Prerequisites:
Logarithm (definition)
Product of the powers property (proof)

Proof:
Let x = a^z,\; where a is the base of the log. Then by definition of log,

\qquad\quad log_a{(x)} = z\qquad\qquad\qquad \qquad\;\;\;\cdots (1)

\begin{align}\text{Now, }\; log_a{(x^y)} &= log_a{\big(a^z\big)^y}\\ &= log_a{(a^{yz})}\qquad\qquad\quad\text{(by product of the power property)}\\ &= yz\qquad\qquad\qquad\qquad\text{(by definition of log)}\\ &= y\; log_a{(x)}\qquad\qquad\quad\text{(from $(1)$)}\end{align}

Hence the result.

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