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

We want to prove:

    alogb(c) = clogb(a).

Recall that we can always write x = eln(x), so xy = (eln(x))y = ey ln(x).

Recall also the change of basis formula for logs:

logb (x) = y  <=>  by = x  <=>  y ln(b) = ln(x)  <=>  y = logb(x) = ln(x) / ln(b).

Putting these two remarks together, we have:

    alogb(c) = elogb(c) ln(a) = e[ln(c) / ln(b)] ln(a) = e[ln(a) / ln(b)] ln(c) = elogb(c) ln(a) = clogb (a).

Q.E.D.

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