Logarithmic Equation

$\begin{aligned} (x+1) & = 2\log_2(2^x+3)-2\log_4(1980-2^{-x}) \\ & = 2\log_2(2^x+3)-2\times \frac {\log_2(1980-2^{-x})}{\log_2 4} \\ & = 2\log_2(2^x+3) - \log_2 \left(1980-\frac 1{2^x}\right) \\ & = \log_2 \left(\frac {(2^x+3)^2}{1980-\frac 1{2^x}}\right) \end{aligned}$

$\begin{aligned} \implies \frac {2^x(2^x+3)^2}{1980(2^x)-1} & = 2^{x+1} \\ \left(2^x+3\right)^2 & = 2\left(1980(2^x)-1\right) \\ 2^{2x} + 6(2^x) + 9 & = 3960(2^x) - 2 \\ 2^{2x} - 3954(2^x) + 11 & = 0 \end{aligned}$

Let the roots of $x$ be $a$ and $b$ . Then by Vieta's formula , we have $2^a \times 2^b = 2^{a+b} = 11$ $\implies a+b = \log_2 11$ . Therefore, $\alpha + 2\beta = 11+2 \times 2 = \boxed{15}$ .