ln(x-1)^2+ln(x^2-1)+ln(x+1)^2=0
Hjelp meg, takk
naturlige logaritme likning
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John Cena54
- Cantor
- Posts: 123
- Joined: 03/11-2006 19:44
jeg trenger hjelp til denne likningen
ln(x-1)^2+ln(x^2-1)+ln(x+1)^2=0
Hjelp meg, takk
ln(x-1)^2+ln(x^2-1)+ln(x+1)^2=0
Hjelp meg, takk
[tex]ln (x - 1)^2 + ln (x^2 - 1) + ln (x + 1)^2 = 0[/tex]
[tex]ln ( (x-1)^2 \cdot (x^2 -1) \cdot (x+1)^2 )= 0[/tex]
[tex]ln ((x^2-1) \cdot (x-1)^2 \cdot (x+1)^2) = 0[/tex]
[tex]ln ((x^2-1) \cdot (x-1)(x-1)\cdot (x+1)(x+1)) = 0[/tex]
[tex]ln ((x^2-1) \cdot (x-1)(x+1) \cdot (x-1)(x+1)) = 0[/tex]
[tex]ln ((x^2-1) \cdot (x^2-1) \cdot (x^2-1) ) = 0[/tex]
[tex]ln ((x^2-1)^3) = 0[/tex]
[tex]3 \cdot ln (x^2-1) = 0[/tex]
[tex]ln (x^2-1) = 0[/tex]
[tex]e^{ln (x^2-1)} = e^0[/tex]
[tex]x^2 - 1 = 1[/tex]
[tex]\underline{\underline{x = \pm \sqrt{2}}[/tex]
[tex]ln ( (x-1)^2 \cdot (x^2 -1) \cdot (x+1)^2 )= 0[/tex]
[tex]ln ((x^2-1) \cdot (x-1)^2 \cdot (x+1)^2) = 0[/tex]
[tex]ln ((x^2-1) \cdot (x-1)(x-1)\cdot (x+1)(x+1)) = 0[/tex]
[tex]ln ((x^2-1) \cdot (x-1)(x+1) \cdot (x-1)(x+1)) = 0[/tex]
[tex]ln ((x^2-1) \cdot (x^2-1) \cdot (x^2-1) ) = 0[/tex]
[tex]ln ((x^2-1)^3) = 0[/tex]
[tex]3 \cdot ln (x^2-1) = 0[/tex]
[tex]ln (x^2-1) = 0[/tex]
[tex]e^{ln (x^2-1)} = e^0[/tex]
[tex]x^2 - 1 = 1[/tex]
[tex]\underline{\underline{x = \pm \sqrt{2}}[/tex]
Last edited by ettam on 12/05-2007 21:02, edited 1 time in total.
Jeg regner med du mener ln a[sup]2[/sup], og ikke (ln a)[sup]2[/sup]. Husk at ln a + ln b = ln ab
[tex]\ln (x-1)^2 + \ln(x^2-1) + \ln(x+1)^2 = 0 \\ \ln (x-1)^2(x+1)^2 + \ln(x^2-1) = 0 \\ \ln (x^2-1)^2 + \ln (x^2-1) = 0 \\ 2\ln(x^2-1) + \ln (x^2-1) = 0 \\ 3\ln(x^2-1) = 0[/tex]
Du tar det derfra
Edit: Ettam kom meg i forkjøpet.
[tex]\ln (x-1)^2 + \ln(x^2-1) + \ln(x+1)^2 = 0 \\ \ln (x-1)^2(x+1)^2 + \ln(x^2-1) = 0 \\ \ln (x^2-1)^2 + \ln (x^2-1) = 0 \\ 2\ln(x^2-1) + \ln (x^2-1) = 0 \\ 3\ln(x^2-1) = 0[/tex]
Du tar det derfra
Edit: Ettam kom meg i forkjøpet.