Find Value of 'a'

$a^{n}+a^{2n}=a^{3n}$

$\Rightarrow a^{1+\log_2 \cos \frac{\pi}{5}}+a^{2(1+\log_2 \cos \frac{\pi}{5})}=a^{3(1+\log_2 \cos \frac{\pi}{5})}$ (since n=1+\log_2 \frac \cos{\pi}{5}

We have $a^{\log_a b}=b$ (easy to prove this) so that we have

$a^{2} \cos^{2} \frac{\pi}{5}-a\cos \frac{\pi}{5}-1=0$

This is a quadratic equation, since $a>0$ because $a$ base of a logarithm so that we have the result $a=2$