2011/110) solve for a If a^n + a^(2n) = a^(3n) where n = 1+ log(base a)(cos(π/5)

a cannot be zero as base to a is taken

For a non trivial solution

a is not zero and dividing both sides by a^n we get

1+ a^n = a^2n

So a^2n = golden ration φ

n = 1+ log(base a)(cos(π/5)

so a^n = a * (cos(π/5)

= a * φ/2 = φ

Or a= 2