Prove x^(log_(b)y) = y^(log_(b)x)?

Question

fyi: b is the base of the log

Answer 1

Well, let t = log_(b)y, hence b^t = y
The left side of the equation is now: x^(log_(b)y) = x^t
The right side of the equation is now:
y^(log_(b)x)
=(b^t)^(log_(b)x)
=b^[(log_(b)y)*(log_(b)x)]
=b^[(log_(b)x)*(log_(b)y)]
=[b^(log_(b)x)]^(log_(b)y)
=x^t <>
Therefore x^(log_(b)y) = y^(log_(b)x)