Let $2^x=3x$. Show that the equation has a solution in the range $(0, 1).$

Let $2^x=3x$. Show that the equation has a solution in the range $(0, 1).$

I was trying to apply just Bolzano’s theorem here, but seems that that’s not quite enough?

Denoting $f(x) = 2^x-3x$ and differentiating one has $f'(x) = 2^{x-1}x -3$ and then picking for example $x=frac32$ I have $f'(frac{3}{2})=frac{3cdot :2^{frac{1}{2}}}{2}-3 <0$ and $x=frac{1}{2}$ I have $f'(frac{1}{2})=frac{sqrt{2}}{4}-3<0$ which are both unfortunately negative…

Looking at the graph of the derivative it seems that it’s negative everywhere in $(0,1).$ I guess I just have to use limit definition or something here?