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.