Q; For what value of n does 3^1998 + 9^999 + 27^n = 3^1999?

(^ = to the power of)

A: 666

Rewriting each number in base 3:

3^1998 + (3^2)^999 + (3^3)^n = 3^1999

3^1998 + 3^1998 + 3^(3n) = 3^1999

Rewriting each number in base 3:

3^1998 + (3^2)^999 + (3^3)^n = 3^1999

3^1998 + 3^1998 + 3^(3n) = 3^1999

It takes 3 3^1998s to make 3^1999 as 3 x 3^1998 = 3^1999

Therefore 3^(3n) = 3^1998

so 3n = 1998 and n = 666