Show that if a<1 that log_a(x) = -log_(1/a)(x).?

Answer 1

Say that y = log_a(x). Then by definition

x = a^y (where ^ means exponent)

x = 1 / (1/a^y)

x = 1 / (1/a)^y

x = (1/a)^(-y) (negative exponent --> inverse of positive exponent)

This translates into log_(1/a)(x) = -y. Thus y = -log_(1/a)(x). QED.

Answer 2

For every a>0, not just for 00. Remember that log_a(1) =0

Answer 3

if a<1 that log_a(x) = -log_(1/a)(x)
left side of equation:
log_a (x) =y is the same as a^y=x

right side:
log_(1/a)(x)=w is the same as (1/a)^w =x
a^{-w} =x
so y=-w
i.e.
log_a (x) = - log_(1/a)(x)