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

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*}

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