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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