Show that, for all a, b and c, a^log_b (c) = c^log_b (a).

We want to prove:

    a^log_b(c) = c^log_b(a).

Recall that we can always write x = e^ln(x), so x^y = (e^ln(x))^y = e^{y ln(x)}.

Recall also the change of basis formula for logs:

log_b (x) = y  <=>  b^y = x  <=>  y ln(b) = ln(x)  <=>  y = log_b(x) = ln(x) / ln(b).

Putting these two remarks together, we have:

    a^log_b(c) = e^{log_b(c) ln(a)} = e^{[ln(c) / ln(b)] ln(a)} = e^{[ln(a) / ln(b)] ln(c)} = e^{log_b(c) ln(a)} = c^{log_b (a)}.

Q.E.D.