## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$1.262$
Using change-of-base, $\displaystyle \log_{b}M=\frac{\log_{a}M}{\log_{a}b}$, we change base to 10 $\displaystyle \log_{9}16=\frac{\log 16}{\log 9}$ ... next, we recognize $16=2^{4},\ 9=3^{2}$ ... and apply $\log_{a}M^{p}=p\cdot\log_{a}M$. $... =\displaystyle \frac{4\log 2}{2\log 3}$ ... substitute $\log 2=0.301$ and $\log 3=0.477,$ $=\displaystyle \frac{4\times 0.301}{2\times 0.477}\approx$1.26205450734 To three decimal places, $=1.262$