1, log0,5(2x + 1), 25 - mają tworzyć ciąg geometryczny, więc musi on mieć postać 1,5,25 lub 1,-5,25. ---------------------------- log0,5(2x+1)=5 v log0,5(2x+1)=-5 czyli mamy: [latex](0,5)^{5} = 2x+1[/latex] v [latex](0,5)^{-5} = 2x+1[/latex] [latex](frac{1}{2})^{5} = 2x+1[/latex] v [latex](frac{1}{2})^{-5} = 2x+1[/latex] [latex]frac{1}{32}= 2x+1[/latex] v [latex]32= 2x+1[/latex] [latex]-frac{31}{32}= 2x[/latex] v [latex]31= 2x[/latex] [latex]-frac{31}{64}= x[/latex] v [latex]x= frac{31}{2}[/latex]
[latex]\log_{0,5}^2(2x+1)=1*25, zal. 2x+1>0implies x>-frac12 \log_{0,5}(2x+1)=t \t^2-25=0 \(t+5)(t-5)=0 \log_{0,5}(2x+1)=-5 vee log_{0,5}(2x+1)=5 \log_{0,5}(2x+1)=log_{0,5}32 vee log_{0,5}(2x+1)=log_{0,5}frac{1}{32} \2x+1=32 vee 2x+1=frac{1}{32} \2x=31/:2 vee 2x=-frac{31}{32} \x=15,5 vee x=-frac{31}{64}[/latex]