Sum of real roots of exponential equation

Let $y = 10^x > 0$. The equation becomes: $y + \frac{4}{y} = \frac{81}{2} \implies 2y^2 - 81y + 8 = 0$

Product of roots $y_1 y_2 = \frac{8}{2} = 4$. Since sum of roots $> 0$ and product $> 0$, both roots $y_1, y_2$ are positive real numbers.

Since $x_1 = \log_{10} y_1$ and $x_2 = \log_{10} y_2$: $x_1 + x_2 = \log_{10} y_1 + \log_{10} y_2 = \log_{10}(y_1 y_2) = \log_{10} 4 = 2 \log_{10} 2$