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 
Math Confidence by Robin the Math Lady 