Question

Упростите выражение:
$(log_{a}{b}+log_{b}{a}+2)(log_{a}{b}-log_{ab}{b})log_{b}{a}-1$

Answer 1

$(log_{a}{b}+log_{b}{a}+2)(log_{a}{b}-log_{ab}{b})log_{b}{a}-1 = \ = left(log_{a}{b}+ dfrac{1}{log_{a}{b}} +2right) left(log_{a}{b}- dfrac{1}{log_{b}{ab}} right) dfrac{1}{log_{a}{b}} -1 = \ =left(dfrac{log^2_{a}{b}}{log_{a}{b}}+ dfrac{1}{log_{a}{b}} +dfrac{2log_{a}{b}}{log_{a}{b}}right) left(log_{a}{b}- dfrac{1}{log_{b}{a}+log_{b}{b}} right)dfrac{1}{log_{a}{b}}-1 =$
$=dfrac{log^2_{a}{b}+2log_{a}{b}+1}{log_{a}{b}} left(log_{a}{b}- dfrac{1}{log_{b}{a}+1} right)dfrac{1}{log_{a}{b}}-1 = \ =dfrac{(log_{a}{b}+1)^2}{log^2_{a}{b}} left(log_{a}{b}- dfrac{1}{ frac{1}{log_{a}{b}} +1} right)-1 = \ =dfrac{(log_{a}{b}+1)^2}{log^2_{a}{b}} left(log_{a}{b}- dfrac{1}{ frac{1+log_{a}{b}}{log_{a}{b}} } right)-1 = \ =dfrac{(log_{a}{b}+1)^2}{log^2_{a}{b}} left(log_{a}{b}- dfrac{log_{a}{b}}{1+log_{a}{b} } right)-1 =$
$=dfrac{(log_{a}{b}+1)^2}{log^2_{a}{b}} cdot dfrac{log_{a}{b}+log^2_{a}{b}-log_{a}{b}}{1+log_{a}{b} } -1 = \ = dfrac{(log_{a}{b}+1)^2}{log^2_{a}{b}} cdot dfrac{log^2_{a}{b}}{1+log_{a}{b} } -1 =(log_{a}{b}+1)-1=log_{a}{b}$