Exponential Functions
posted by Jus on .
Show that if
log_b (a) = c, and log_y (b) = c, then log_a (y)=c^2

recall
log_{b} a = log a/log b (base 10, or any other base for that matter)
so log_{b}(a) = loga/logb = c
and log_{y}(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 = log_{a}(y)
so log_{a}(y) = c^2 as required.