Posted by Jus on Tuesday, May 19, 2009 at 7:51pm.
recall
logb a = log a/log b (base 10, or any other base for that matter)
so logb(a) = loga/logb = c
and logy(b) = logb/logy = c
then [loga/logb][logb/logy] = (c)(c) = c^2
so loga/logy = c^2 or
logy/loga = 1/c^2 = c^-2
but logy/loga = loga(y)
so loga(y) = c^-2 as required.
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