If aloga(a^a) =16, find 16^a.?

a = a^1

Therefore log (base a) a = 1

(this is true for any value of a : log(base 10) 10 = 1; ln e = 1; and so on)

If you raise a number to a power, then its logarithm is multiplied by that power, so

log a^a = 1 x a

and a log a^a = a (1 x a) = a^2

So you can look at alog(base a) (a^a) as (a^2)log a, ..... ..... (1)

or as log a^(a^2) ..... ..... ..... ..... (2)

(1) makes the equation (a^2) x 1 = 16

(2) makes the equation a^2 = 16 (but arrived at by a slightly different route - I thought I would give you both)

In either case, we have a^2 = 16, so a = 4

You should be able to finish it off now.

alog_a(a^a) = 16
a(a)log_a(a) = 16
a(a)(1) = 16
a(a) = 16
a^2 = 16
a = ±√16
a = ±4

16^a
= 16^4
= 65536

16^a
= 16^(-4)
= 1/16^4
= 1/65536