How to solve this? #(1/2)^x=x^(1/2)#.

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Answer 1 · 2017-05-26T20:30:47

#x=1/2#

One solution is #x=1/2#, since it makes both sides of the equation the same, namely:

#(1/2)^(1/2)#

Are there other (real) solutions?

Let's draw a graph of #y=(1/2)^x# and #y=x^(1/2)#:

graph{(y-(1/2)^x)(y^2-x)(sqrt(y+0.02)) = 0 [-6.04, 13.96, -4.52, 5.48]}

It seems they will only cross at one point.

Note that: #(1/2)^x# is monotonically decreasing and #x^(1/2)# is monotonically increasing.