Let a,b be real #'s - {0}. Prove that log_a(x) = log_b(x)/log_b(a) by showing that a^(log_b(x)/log_b(a)) = x.

Answer 1

log_a (x)=y is the number to which you need to raise a to get x, so
a^y=x.

on the other hand
log_b (x)=z is the number to which you need to raise b to get x. so
b^z=x
and finally log_b a=w is the number to which you need to raise b to get a:
b^w=a.

Answer 2

You actually need both a and b to be positive rather than simply non-zero. Use the fact that a=b^(log_b (a)). Raise both sides of this to the power of your exponent and a miracle happens.

Answer 3

let be a = b^t

a^(log_b(x)/log_b(a)) =
= (b^t)^(log_b(x)/log_b(b^t)) =
= b^(t(log_b(x)/t)) =
= b^(log_b(x)) =
= x

thus

a^(log_b(x)/log_b(a)) = x = a^(log_a(x))

therefore log_b(x)/log_b(a) = log_a(x)

Answer 4

The question is not clear. give more information.