Prove that log (a^r) = r log a for all rationals r, for a>0

by RoRi

November 6, 2015

If $a> 0$ and $r \in \mathbb{Q}$ , prove that

$\log \left( a^r \right) = r \log a.$

Proof. Since $r \in \mathbb{Q}$ we may write $r = \frac{m}{n}$ for some $m,n \in \mathbb{Z}$ . Therefore,

$\begin{align*} \log \left(a^r \right) &= \log \left( a^{\frac{m}{n}} \right) \\ &= \log \left(\left( a^\frac{1}{n} \right)^m\right) \\ &= m \log a^{\frac{1}{n}} &(\text{Funct. Eqn of } log) \\ &= \left( \frac{m}{n} \right) n \log a^{\frac{1}{n}} \\ &= \frac{m}{n} \left( \left( a^{\frac{1}{n}} \right)^n \right) &(\text{Func. Eqn})\\ &= \frac{m}{n} \log a \\ &= r \log a. \qquad \blacksquare \end{align*}$

Stumbling Robot

Prove that log (a^r) = r log a for all rationals r, for a>0

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply