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Mathematics LibreTexts

8.4: The Logarithm Functions

  • Page ID
    22688
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    Definition

    Given a positive real number \(x,\) we call

    \[\log (x)=\int_{1}^{x} \frac{1}{t} d t\]

    the logarithm of \(x .\)

    Note that \(\log (1)=0, \log (x)<0\) when \(0<x<1,\) and \(\log (x)>0\) when \(x>1 .\)

    Proposition \(\PageIndex{1}\)

    The function \(f(x)=\log (x)\) is an increasing, differentiable function with

    \[f^{\prime}(x)=\frac{1}{x}\]

    for all \(x>0 .\)

    Proof

    Using the Fundamental Theorem of Calculus, we have

    \[f^{\prime}(x)=\frac{1}{x}>0\]

    for all \(x>0,\) from which the result follows. \(\quad\) Q.E.D.

    Proposition \(\PageIndex{2}\)

    For any \(x>0\),

    \[\log \left(\frac{1}{x}\right)=-\log (x).\]

    Proof

    Using the substitution \(t=\frac{1}{u},\) we have

    \[\log \left(\frac{1}{x}\right)=\int_{1}^{\frac{1}{x}} \frac{1}{t} d t=\int_{1}^{x} u\left(-\frac{1}{u^{2}}\right) d u=-\int_{1}^{x} \frac{1}{u} d u=-\log (x).\]

    Q.E.D.

    Proposition \(\PageIndex{3}\)

    For any positive real numbers \(x\) and \(y,\)

    \[\log (x y)=\log (x)+\log (y).\]

    Proof

    Using the substitution \(t=x u,\) we have

    \[\begin{aligned} \log (x y) &=\int_{1}^{x y} \frac{1}{t} d t \\ &=\int_{\frac{1}{x}}^{y} \frac{x}{x u} d u \\ &=\int_{\frac{1}{x}}^{1} \frac{1}{u} d u+\int_{1}^{y} \frac{1}{u} d u \\ &=-\int_{1}^{\frac{1}{x}} \frac{1}{u} d u+\log (y) \\ &=-\log \left(\frac{1}{x}\right)+\log (y) \\ &=\log (x)+\log (y). \end{aligned}\]

    Q.E.D.

    Proposition \(\PageIndex{4}\)

    If \(r \in \mathbb{Q}\) and \(x\) is a positive real number, then

    \[\log \left(x^{r}\right)=r \log (x).\]

    Proof

    Using the substitution \(t=u^{r},\) we have

    \[\log \left(x^{r}\right)=\int_{1}^{x^{r}} \frac{1}{t} d t=\int_{1}^{x} \frac{r u^{r-1}}{u^{r}} d u=r \int_{1}^{x} \frac{1}{u} d u=r \log (x).\]

    Q.E.D.

    Proposition \(\PageIndex{5}\)

    \(\lim _{x \rightarrow+\infty} \log (x)=+\infty\) and \(\lim _{x \rightarrow 0+} \log (x)=-\infty .\)

    Proof

    Given a real number \(M,\) choose an integer \(n\) for which \(n \log (2)>M\) (there exists such an \(n\) since \(\log (2)>0\) ). Then for any \(x>2^{n}\), we have

    \[\log (x)>\log \left(2^{n}\right)=n \log (2)>M.\]

    Hence \(\lim _{x \rightarrow+\infty} \log (x)=+\infty\).

    Similarly, given any real number \(M,\) we may choose an integer \(n\) for which \(-n \log (2)<M .\) Then for any \(0<x<\frac{1}{2^{n}},\) we have

    \[\log (x)<\log \left(\frac{1}{2^{n}}\right)=-n \log (2)<M.\]

    Hence \(\lim _{x \rightarrow 0+} \log (x)=-\infty . \quad\) Q.E.D.

    Note that the logarithm function has domain \((0,+\infty)\) and range \((-\infty,+\infty)\).

    Exercise \(\PageIndex{1}\)

    Show that for any rational number \(\alpha>0\),

    \[\lim _{x \rightarrow+\infty} x^{\alpha}=+\infty .\]

    Proposition \(\PageIndex{6}\)

    For any rational number \(\alpha>0\),

    \[\lim _{x \rightarrow+\infty} \frac{\log (x)}{x^{\alpha}}=0.\]

    Proof

    Choose a rational number \(\beta\) such that \(0<\beta<\alpha .\) Now for any \(t>1\),

    \[\frac{1}{t}<\frac{1}{t} t^{\beta}=\frac{1}{t^{1-\beta}}.\]

    Hence

    \[\log (x)=\int_{1}^{x} \frac{1}{t} d t<\int_{1}^{x} \frac{1}{t^{1-\beta}} d t=\frac{x^{\beta}-1}{\beta}<\frac{x^{\beta}}{\beta}\]

    whenever \(x>1 .\) Thus

    \[0<\frac{\log (x)}{x^{\alpha}}<\frac{1}{\beta x^{\alpha-\beta}}\]

    for \(x>1 .\) But

    \[\lim _{x \rightarrow+\infty} \frac{1}{\beta x^{\alpha-\beta}}=0,\]

    so

    \[\lim _{x \rightarrow+\infty} \frac{\log (x)}{x^{\alpha}}=0.\]

    Q.E.D.

    Exercise \(\PageIndex{2}\)

    Show that

    \[\lim _{x \rightarrow 0^{+}} x^{\alpha} \log (x)=0\]

    for any rational number \(\alpha>0\).

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