Correct.

$\displaystyle{2}={\log}_{{{b}}}{64}\usepackage{cmbright}\Huge\$ is equivalent to $\displaystyle{{b}}^{{2}}={64}\usepackage{cmbright}\Huge\$ .

$\displaystyle{{2}}^{{2}}={4}\usepackage{cmbright}\Huge\$ and $\displaystyle{{10}}^{{2}}={100}\usepackage{cmbright}\Huge\$ , so b is between 2 and 10.

Now, $\displaystyle{{8}}^{{2}}={8}\cdot{8}={64}\usepackage{cmbright}\Huge\$ , so $\displaystyle{b}={8}\usepackage{cmbright}\Huge\$ .