Hello,

How do I solve the following equation? $\displaystyle 2^2^x-3*2^x-40 = 0$

I tried the following, but apparently this is not correct, i.e. score = 0%.

$\displaystyle 2^2^x-3*2^x-40 = 0$

$\displaystyle 2^2^x-\sqrt{3}*2^2^x = 40$

$\displaystyle (1-\sqrt{3})*2^2^x = 40$

$\displaystyle 2^2^x = \frac{40}{1-\sqrt{3}}$

$\displaystyle 2x = \log_{2}\frac{40}{1-\sqrt{3}}$

$\displaystyle x = \frac{\log_{2}\frac{40}{1-\sqrt{3}}}{2}$