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December 17, 2014

December 17, 2014

Posted by **Jus** on Tuesday, May 19, 2009 at 7:51pm.

log_b (a) = c, and log_y (b) = c, then log_a (y)=c^-2

- Exponential Functions -
**Reiny**, Tuesday, May 19, 2009 at 8:31pmrecall

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.

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