Solution: Let us suppose that the given equation has more than three solutions, $10,1,3$ and $\alpha \in \mathbb{R} $. Consider the function \[ f(x) = 2^x - \frac{2}{3}x^2 - \frac{1}{3}x - 1. \] Since $f(0) = f(1) = f(3) = f(\alpha ) = 0$, by Rolle's theorem there exists real numbers $x_1 \lt x_2 \lt x_3 $ such that $f^\prime \left( x_i \right) = 0$ for $i = 1,2,3$. Now again use the Rolle's theorem to find real numbers $y_1 \lt y_2$ such that $f^{\prime\prime} \left( y_1 \right) = 0 f^{\prime\prime} \left( y_2 \right) $. We will apply the Rolle's theorem one more time to find a real number $z$ such that $f^{\prime \prime \prime} (z) = 0$. But observe that \begin{align*} f(x) = 2^x - \frac{2}{3}x^2 - \frac{1}{3}x - 1 & \implies f^\prime (x) = 2^x \ln 2 - \frac{4}{3}x - \frac{1}{3} \\ & \implies f^{\prime\prime} (x) = 2^x (\ln 2)^2 - \frac{4}{3} \\ & \implies f^{\prime\prime \prime} (x) = 2^x (\ln 2)^3 > 0, \end{align*} hence we arrived at a contradiction. Therefore, the given equation has only three solutions.