Hvordan løser man denne logaritmeoppgaven
lgx[sup]2[/sup]-lg([tex]\frac{x}{2}[/tex]+1)=1
Logaritme
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]lg x^2 - lg (\frac{x}{2} + 1) = 1[/tex]
[tex]lg (\frac{x^2}{\frac {x}{2} + 1}) = 1[/tex]
[tex]10^{lg (\frac{x^2}{\frac {x}{2} + 1})} = 10^1[/tex]
[tex]\frac{x^2}{\frac {x}{2} + 1} = 10[/tex]
[tex]x^2 = 10(\frac {x}{2} + 1)[/tex]
[tex]x^2 = 5x + 10[/tex]
[tex]x^2 - 5x - 10 = 0[/tex]
Resten løser du selv...
[tex]lg (\frac{x^2}{\frac {x}{2} + 1}) = 1[/tex]
[tex]10^{lg (\frac{x^2}{\frac {x}{2} + 1})} = 10^1[/tex]
[tex]\frac{x^2}{\frac {x}{2} + 1} = 10[/tex]
[tex]x^2 = 10(\frac {x}{2} + 1)[/tex]
[tex]x^2 = 5x + 10[/tex]
[tex]x^2 - 5x - 10 = 0[/tex]
Resten løser du selv...