F(x)=log base3 of x, then f(32)=?

Question

its not base 3 its base 2

Answer 1

log_2 32 = log_2 2^5 = 5

That's all

Answer 2

f(32) = log base 2 (32) = log base 2(2^5)

Since log base 2(2) = 1,

f(32) = log base 2(2)+log base 2(2)+log base 2(2)+log base 2(2)+log base 2(2) = 5