[nooonuii]

Define $f(x)=2^x-\dfrac{3x}{2}$ for $x\geq0$.

Then $f'(x)=2^x\ln{2}-3/2$.

$f''(x)=2^x(\ln{2})^2>0$

$f'(x)=0$ iff $x=\log_2\Big({\dfrac{3}{\ln{4}}}\Big)$.

Thus $f$ has a minimum at $x_0=\log_2\Big({\dfrac{3}{\ln{4}}}\Big)$.

But $f(x_0)>0$.

This means $f(x)>0$ for all $x\geq 0$.

If $x$ is the solution of $2^x=\dfrac{3x}{2}$ then $x=\dfrac{2}{3} 2^x>0$.

Thus $f(x)=0$ which is a contradiction.

Therefore the equation $2^x=\dfrac{3x}{2}$ has no solution.