# algebra

Writ the equation in Logarithm form
25=(1/5)^2
a=b^c

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1. Logb(a) = c

Log1/5(25) = 2

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posted by Kuai
2. but 25 is not (1/5)^2
(1/5)^2 = 1/25
25 = 1/(1/5)^2 = (1/5)^-2
So, log1/525 = -2

loga = c*logb
regardless of base. Using b for the base, log_b(b)=1,so
logba = c

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posted by Steve

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