# How do you evaluate log_2 1?

${\log}_{2} 1 = 0$
From the definition of logarithm, if ${a}^{m} = b$ then ${\log}_{a} b = m$.
Now as for all values of $a$, ${a}^{0} = 1$, hence ${\log}_{a} 1 = 0$
Hence ${\log}_{2} 1 = 0$