Naturlige logarimer - grunntall $e$

\begin{align}\ln e&=1\\e^{\ln a}&=a\\\ln e^{a}&=a\\\ln a^b&=b\cdot \ln a\\\ln (a\cdot b)&=\ln a + \ln b\\\ln \Big(\frac{a}{b}\Big)&=\ln a - \ln b\end{align}

Oppgave 1

  • Løsningsforslag:

    $\ln e^x=x\cdot \ln e=x$

  • Løsningsforslag:

    $2\ln e^2=2\cdot \ln e = 2$

  • Løsningsforslag:

    $\ln e^{-3}=-3$

  • Løsningsforslag:

    $\ln(e^4\cdot e^2) =\lne^4+\lne^2=4+2=6$

    eller

    $\ln(e^4\cdot e^2) =\ln e^{4+2}=\ln e^6=6$

  • Løsningsforslag:

    $\ln(\sqrt[3]{e^2})=\ln e^{2/3}=\frac{2}{3}$

Oppgave 2

  • Løsningsforslag:

    \begin{align*}\ln x^5 - \ln \frac{1}{x^3}-2\ln x^4 &=5\ln x-(\ln1-3\ln x)-8\ln x\\
    &=5\ln x+3\ln x-8\ln x\\
    &=0
    \end{align*}

  • Løsningsforslag:

    \begin{align*}
    \ln(x^2 y)-\ln \frac{x}{y^2}-\ln \frac{x^2}{y}&= 2\ln x+\ln y-(\ln x-2\ln y)-(2\ln x-\ln y)\\
    &=2\ln x+\ln y-\ln x+2\ln y-2\ln x+\ln y\\
    &=-\ln x+4\ln y
    \end{align*}

  • \begin{align}
    \ln\frac{3}{y}+\ln (9y^3)-\ln 27
    &=\ln 3-\ln y+\ln 3^2+\ln y^3-\ln 3^3\\
    &=\ln 3-\ln y+2\ln 3+3\ln y-3\ln 3\\
    &=2\ln y
    \end{align}

  • \begin{align*}
    \ln 16 + \ln 4x^2 - 2\ln 2 &=\ln 2^4 + \ln2^2+\ln x^2-2\ln 2\\
    &=4\ln 2+2\ln 2+2\ln x-2\ln 2\\
    &=2\ln x+4\ln 2
    \end{align*}

  • \begin{align*}
    2\ln b-\ln\frac{1}{b}-\ln(ab^2)+\ln\frac{a}{b^2}
    &=2\ln b-\ln 1+\ln b-\ln a-2\ln b+\ln a-2\ln b\\
    &=-\ln b
    \end{align*}

Oppgave 3

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Oppgave 4

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Oppgave 5

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Oppgave 6

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