## Precalculus (10th Edition)

Since $\log_a {x^n}=n\cdot \log_a {x}$, then $f(x^{\alpha})=\log_a{(x^{\alpha})}=\alpha \cdot \log_a{x}=\alpha \cdot f(x)$
We know that $\log_a {x^n}=n\cdot \log_a {x}$. Hence, $f(x^{\alpha})=\log_a{x^{\alpha}}=\alpha \cdot \log_a{x}=\alpha \cdot f(x)$