## Functions Modeling Change: A Preparation for Calculus, 5th Edition

$\log 25 \approx 1.4$
Use the fact that $(a^b)^n=a^{bn}$ to simplify the expression to $25=5^2=(10^{0.7})^2=10^{1.4}$. Therefore, $$\log 25=\log 10^{1.4}$$ Since $\log 10^a=a$ for all positive $a$, $\log 10^{1.4}\approx1.4$