We start with #-5=log_2x#. Now, we can rewrite a logarithm as an exponentiation equation, like this: #color(red)(y)=color(blue)(b)^(color(green)(x)# becomes #log_color(blue)(b)color(red)(y)=color(green)(x)# and vice versa. So, #log_color(blue)(2)color(red)(x)=color(green)(-5)# becomes #color(red)(x)=color(blue)(2)^(color(green)(-5)#.
And if we solve #2^-5# we get #x=1/32# or #.03125#.