Question 1058942

I solved by the following reasoning:

Slope of log2(x) over (1,2) is  positive and rising from 0 to 1  (  log2(1)=0, log2(2)=1  )

Slope of {{{ 2^(-x)  = 1/2^x }}}  is negative and the function decreases over (1,2) from 1/2 to 1/4 


So log2(1) < 1/2  and log2(2) > 1/4 

That means the two functions cross on (1,2) and since they have slopes of opposite signs, they will only cross once on that interval.