Whether we use the principal root or not, the function on the $LHS$ is symmetric around $x=0,$ where it gets its maximum value of $2$. So, for a single real root, we have $a=2.$

tk1 wrote: Whether we use the principal root or not, the function on the $LHS$ is symmetric around $x=0,$ where it gets its maximum value of $2$. So, for a single real root, we have $a=2.$ Not exactly. What you proved is that if $a\ne 2$, then $x=0$ is not a solution and so we have an even number of solutions. So, in order to have an odd number of solutions, we need $a=2$ but maybe we have $3$ roots then. So $a=2$ is a necessary condition, but maybe not sufficient. So you miss, according to me, a last step : show that when $a=2$ we indeed have only $x=0$ as solution.