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f(x) = 2 lg(x^2 - 2x) + 3

 26.07.2020
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Nullstelle finden:

 

 \(\color{BrickRed}f(x) = 2\cdot lg(x^2 - 2x) + 3=0\\ lg(x^2-2x)=-1,5\\ x^2-2x=10^{-1,5}\\ x^2-2x-0,03162=0 \)

\(x=1\pm\sqrt{1+10^{-1,5}}\\ x=1\pm 1,0157\)

\(\color{blue}x_1=2,0157\\ x_2=-0,0157\ \small{entfaellt,\ lg\ ist\ komplex}\)

\(\mathbb L=\{2,0157\}\)

 

Probe:  \(2\cdot lg(2,0157^2-2\cdot 2,0157)+3\approx 0\)

laugh  !

 26.07.2020
bearbeitet von asinus  26.07.2020

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