wykaz ze -loga b=1/log1/b a

[latex]-log_ab=frac{1}{log_^frac{1}{b}}a}\L=-log_ab=log_ab^{-1}=frac{log_{frac{1}{b}}b^{-1}}{log_{frac{1}{b}}a}=frac{log_{frac{1}{b}}frac{1}{b}}{log_{frac{1}{b}}a}=frac{1}{log_{frac{1}{b}}a}=P\L=P[/latex]