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

$(h)$
The logarithm, base $a$, of 1 is 0$:\qquad \log_{a} 1=0.$ Or, $m=\log_{a}x$ is equivalent to $a^{m}=x.$ so $2^{0}=1 \Leftrightarrow \log_{2}1=0$ case $(h)$