Indices

$ab=a^b$ (for $a\neq 0$ )implies $b=a^{b-1}$ , hence $a=b^{\frac1{b-1}}$ . This expression is not defined for all $b$ , but certainly for all $b>1$ . This gives us $\boxed{\text{infinitely many}}$ ordered pairs $(a,b)\in\mathbb{R}^2$ .

Aditya Raut helped me to get the solution. $a^b=ab$

or $a^(b-1)=b$

or $a=b^{\frac1{b-1}}$

As $a,b$ belongs to real they have infinite set of solutions.