Cao Yi

Logarithm Change of Base Rule

To prove

$ (a \gt 0, a\ne1, b \gt 0, c \gt 0, c \ne 1)(\log_{a}{b} = \dfrac{\log_{c}{b}}{\log_{c}{a}}) $

Proof I

Assume $\log_{a}{b} = x$

$\therefore a^x = b$

$\therefore \log_{c}{a^x} = \log_{c}{b}$

$\therefore x\log_{c}{a} = \log_{c}{b}$

$\therefore x = \dfrac{\log_{c}{b}}{\log_{c}{a}}$

$\therefore \log_{a}{b} = \dfrac{\log_{c}{b}}{\log_{c}{a}}$

Q.E.D.

Proof II

$\because b=a^{\log_{a}{b}}=(c^{\log_{c}{a}})^{\log_{a}{b}}=c^{\log_{c}{a}\log_{a}{b}}$

$\therefore \log_{c}{b}=\log_{c}{a}\log_{a}{b}$

$\therefore \log_{a}{b} = \dfrac{\log_{c}{b}}{\log_{c}{a}}$

Q.E.D.

Proof III

$\because b=c^{\log_{c}{b}}=c^{\log_{c}{a^{\log_{a}{b}}}}=c^{\log_{a}{b}\log_{c}{a}}$

$\therefore \log_{c}{b}=\log_{a}{b}\log_{c}{a}$

$\therefore \log_{a}{b} = \dfrac{\log_{c}{b}}{\log_{c}{a}}$

Q.E.D.

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