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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 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.