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Mathematics LibreTexts

7.4: Properties of the Logarithm

  • Page ID
    6278
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    Skills to Develop

    • Apply the inverse properties of the logarithm.
    • Expand logarithms using the product, quotient, and power rule for logarithms.
    • Combine logarithms into a single logarithm with coefficient 1.

    Logarithms and Their Inverse Properties

    Recall the definition of the base-\(b\) logarithm: given \(b > 0\) where \(b ≠ 1\),

    \(y=\log _{b} x\) if and only if \(x=b^{y}\)

    Use this definition to convert logarithms to exponential form. Doing this, we can derive a few properties:

    \(\begin{aligned}\log _{b} 1=0 &\text { because } b^{0}=1 \\ \log _{b} b=1 & \text { because } b^{1}=b \\ \log _{b}\left(\frac{1}{b}\right)=-1 &\text { because } b^{-1}=\frac{1}{b}\end{aligned}\)

    Example \(\PageIndex{1}\)

    Evaluate:

    1. \(\log 1\)
    2. \(\ln e\)
    3. \(\log _{5}\left(\frac{1}{5}\right)\)

    Solution

    1. When the base is not written, it is assumed to be \(10\). This is the common logarithm, \[\log 1=\log _{10} 1=0 \nonumber\]
    2. The natural logarithm, by definition, has base \(e\), \[\ln e=\log _{e} e=1 \nonumber\]
    3. Because \(5^{-1}=\frac{1}{5}\) we have, \[\log _{5}\left(\frac{1}{5}\right)=-1 \nonumber\]

    Furthermore, consider fractional bases of the form \(1/b\) where \(b > 1\).

    \(\log _{1 / b} b=-1 \quad\) because \(\left(\frac{1}{b}\right)^{-1}=\frac{1^{-1}}{b^{-1}}=\frac{b}{1}=b\)

    Example \(\PageIndex{2}\)

    Evaluate:

    1. \(\log _{1 / 4} 4\)
    2. \(\log _{2 / 3}\left(\frac{3}{2}\right)\)

    Solution

    1. \(\log _{1 / 4} 4=-1 \quad\) because \(\quad\left(\frac{1}{4}\right)^{-1}=4\)
    2. \(\log _{2 / 3}\left(\frac{3}{2}\right)=-1\) because \(\left(\frac{2}{3}\right)^{-1}=\frac{3}{2}\)

    Given an exponential function defined by \(f (x) = b^{x}\) , where \(b > 0\) and \(b ≠ 1\), its inverse is the base-\(b\) logarithm, \(f^{ −1} (x) = log_{b} x\). And because \(f\left(f^{-1}(x)\right)=x\) and \(f^{-1}(f(x))=x\), we have the following inverse properties of the logarithm11:

    \(f^{-1}(f(x))=\log _{b} b^{x}=x\) and
    \(f\left(f^{-1}(x)\right)=b^{\log _{b} x}=x, x>0\)

    Since \(f^{-1}(x)=\log _{b} x\) has a domain consisting of positive values \((0, \infty)\), the property \(b^{\log _{b} x}=x\) is restricted to values where \(x>0\).

    Example \(\PageIndex{3}\)

    Evaluate

    1. \(\log _{5} 625\)
    2. \(5^{\log _{5} 3}\)
    3. \(e^{\ln 5}\)

    Solution

    Apply the inverse properties of the logarithm.

    1. \(\log _{5} 625=\log _{5} 5^{4}=4\)
    2. \(5^{\log _{5} 3}=3\)
    3. \(e^{\ln 5}=5\)

    In summary, when \(b > 0\) and \(b ≠ 1\), we have the following properties:

    Table 7.4.1
    \(\log _{b} 1=0\) \(\log _{b} b=1\)
    \(\log _{1 / b} b=-1\) \(\log _{b}\left(\frac{1}{b}\right)=-1\)
    \(\log _{b} b^{x}=x\) \(b^{\log _{b} x}=x, x>0\)

    Exercise \(\PageIndex{1}\)

    Evaluate: \(\log 0.00001\)

    Answer

    \(-5\)

    http://www.youtube.com/v/YFsZxKiMCTg

    Product, Quotient, and Power Properties of Logarithms

    In this section, three very important properties of the logarithm are developed. These properties will allow us to expand our ability to solve many more equations. We begin by assigning \(u\) and \(v\) to the following logarithms and then write them in exponential form:

    \(\begin{aligned}\log _{b} x=u\color{Cerulean}{\Longrightarrow}\color{black}{b}^{u}=x \\ \log _{b} y=v\color{Cerulean}{\Longrightarrow}\color{black}{b}^{v}=y\end{aligned}\)

    Substitute \(x = b^{u}\) and \(y = b^{v}\) into the logarithm of a product \(log_{b} (xy)\) and the logarithm of a quotient \(\log _{b}\left(\frac{x}{y}\right)\).Then simplify using the rules of exponents and the inverse properties of the logarithm.

    Table 7.4.2
    Logarithm of a Product Logarithm of a Quotient
    \(\begin{aligned} \log _{b}(x y) &=\log _{b}\left(b^{u} b^{v}\right) \\ &=\log _{b} b^{u+v} \\ &=u+v \\ &=\log _{b} x+\log _{b} y \end{aligned}\) \(\begin{aligned} \log _{b}\left(\frac{x}{y}\right) &=\log _{b}\left(\frac{b^{u}}{b^{v}}\right) \\ &=\log _{b} b^{u-v} \\ &=u-v \\ &=\log _{b} x-\log _{b} y \end{aligned}\)

    This gives us two essential properties: the product property of logarithms12,

    \(\log _{b}(x y)=\log _{b} x+\log _{b} y\)

    and the quotient property of logarithms13,

    \(\log _{b}\left(\frac{x}{y}\right)=\log _{b} x-\log _{b} y\)

    In words, the logarithm of a product is equal to the sum of the logarithm of the factors. Similarly, the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.

    Example \(\PageIndex{4}\)

    Write as a sum \(\log _{2}(8 x)\).

    Solution

    Apply the product property of logarithms and then simplify.

    \(\begin{aligned} \log _{2}(8 x) &=\log _{2} 8+\log _{2} x \\[4pt] &=\log _{2} 2^{3}+\log _{2} x \\[4pt] &=3+\log _{2} x \end{aligned}\)

    Answer

    \[3+\log _{2} x \nonumber\]

    Example \(\PageIndex{5}\)

    Write as a difference \(\log \left(\frac{x}{10}\right)\).

    Solution

    Apply the quotient property of logarithms and then simplify.

    \(\begin{aligned} \log \left(\frac{x}{10}\right) &=\log x-\log 10 \\ &=\log x-1 \end{aligned}\)

    Answer

    \[\log x-1 \nonumber\]

    Next we begin with \(log_{b}x = u\) and rewrite it in exponential form. After raising both sides to the \(n\)th power, convert back to logarithmic form, and then back substitute.

    \(\begin{aligned} \log _{b} x&=u \color{Cerulean}{\Longrightarrow}\color{black}{ b}^{u}=x \\&\left(b^{u}\right)^{n} =(x)^{n} \\ \log_{b}x^{n}&=nu\color{Cerulean}{\Longleftarrow}\color{black}{b}^{nu}=x^{n}\\\log_{b}x^{n}&=n\log_{b}x\end{aligned}\)

    This leads us to the power property of logarithms14,

    \(\log _{b} x^{n}=n \log _{b} x\)

    In words, the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.

    Example \(\PageIndex{6}\)

    Write as a product:

    1. \(\log _{2} x^{4}\)
    2. \(\log _{5}(\sqrt{x})\)

    Solution

    1. Apply the power property of logarithms.\[\log _{2} x^{4}=4 \log _{2} x\]
    2. Recall that a square root can be expressed using rational exponents, \(\sqrt{x}=x^{1 / 2}\). Make this replacement and then apply the power property of logarithms. \(\begin{aligned} \log _{5}(\sqrt{x}) &=\log _{5} x^{1 / 2} \\ &=\frac{1}{2} \log _{5} x \end{aligned}\)

    In summary,

    Table 7.4.3
    Product Property of Logarithms \(\log _{b}(x y)=\log _{b} x+\log _{b} y\)
    Quotient Property of Logarithms \(\log _{b}\left(\frac{x}{y}\right)=\log _{b} x-\log _{b} y\)
    Power Property of Logarithms \(\log _{b} x^{n}=n \log _{b} x\)

    We can use these properties to expand logarithms involving products, quotients, and powers using sums, differences and coefficients. A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.

    Caution

    It is important to point out the following:

    \[\log (x y) \neq \log x \cdot \log y\]

    and

    \[\log \left(\frac{x}{y}\right) \neq \frac{\log x}{\log y}\]

    Example \(\PageIndex{7}\)

    Expand completely: \(\ln \left(2 x^{3}\right)\).

    Solution

    Recall that the natural logarithm is a logarithm base \(e\), \(\ln x=\log _{e} x\). Therefore, all of the properties of the logarithm apply.

    \(\begin{aligned} \ln \left(2 x^{3}\right) &=\ln 2+\ln x^{3}\quad \color{Cerulean} { Product\: rule\: for\: logarithms } \\ &=\ln 2+3 \ln x \quad\color{Cerulean} { Power\: rule\: for\: logarithms } \end{aligned}\)

    Answer

    \[\ln 2+3 \ln x \nonumber \]

    Example \(\PageIndex{8}\):

    Expand completely: \(\log \sqrt[3]{10 x y^{2}}\).

    Solution

    Begin by rewriting the cube root using the rational exponent \(\frac{1}{3}\) and then apply the properties of the logarithm.

    \(\begin{aligned} \log \sqrt[3]{10 x y^{2}} &=\log \left(10 x y^{2}\right)^{1 / 3} \\ &=\frac{1}{3} \log \left(10 x y^{2}\right) \\ &=\frac{1}{3}\left(\log 10+\log x+\log y^{2}\right) \\ &=\frac{1}{3}(1+\log x+2 \log y) \\ &=\frac{1}{3}+\frac{1}{3} \log x+\frac{2}{3} \log y \end{aligned}\)

    Answer

    \[\frac{1}{3}+\frac{1}{3} \log x+\frac{2}{3} \log y \nonumber\]

    Example \(\PageIndex{9}\):

    Expand completely: \(\log _{2}\left(\frac{(x+1)^{2}}{5 y}\right)\).

    Solution

    When applying the product property to the denominator, take care to distribute the negative obtained from applying the quotient property.

    \(\begin{aligned} \log _{2}\left(\frac{(x+1)^{2}}{5 y}\right) &=\log _{2}(x+1)^{2}-\log _{2}(5 y) \\ &=\log _{2}(x+1)^{2}-\left(\log _{2} 5+\log _{2} y\right)\quad\color{Cerulean}{Distribute.} \\ &=\log _{2}(x+1)^{2}-\log _{2} 5-\log _{2} y \\ &=2 \log _{2}(x+1)-\log _{2} 5-\log _{2} y \end{aligned}\)

    Answer

    \[2 \log _{2}(x+1)-\log _{2} 5-\log _{2} y \nonumber\]

    Caution

    There is no rule that allows us to expand the logarithm of a sum or difference. In other words,

    \(\log (x \pm y) \neq \log x \pm \log y\)

    Exercise \(\PageIndex{2}\)

    Expand completely: \(\ln \left(\frac{5 y^{4}}{\sqrt{x}}\right)\)

    Answer

    \(\ln 5+4 \ln y-\frac{1}{2} \ln x\)

    http://www.youtube.com/v/h_Zu-SkATl0

    Example \(\PageIndex{10}\):

    Given that \(\log _{2} x=a, \log _{2} y=b\), and that \(\log _{2} z=c\), write the following in terms of \(a\), \(b\), and \(c\):

    1. \(\log _{2}\left(8 x^{2} y\right)\)
    2. \(\log _{2}\left(\frac{2 x^{4}}{\sqrt{z}}\right)\)

    Solution

    1. Begin by expanding using sums and coefficients and then replace \(a\) and \(b\) with the appropriate logarithm. \[\begin{aligned} \log _{2}\left(8 x^{2} y\right) &=\log _{2} 8+\log _{2} x^{2}+\log _{2} y \\ &=\log _{2} 8+2 \log _{2} x+\log _{2} y \\ &=3+2 a+b \end{aligned}\]
    2. Expand and then replace \(a, b\), and \(c\) where appropriate. \[\begin{aligned} \log _{2}\left(\frac{2 x^{4}}{\sqrt{z}}\right) &=\log _{2}\left(2 x^{4}\right)-\log _{2} z^{1 / 2} \\ &=\log _{2} 2+\log _{2} x^{4}-\log _{2} z^{1 / 2} \\ &=\log _{2} 2+4 \log _{2} x-\frac{1}{2} \log _{2} z \\ &=1+4 a-\frac{1}{2} b \end{aligned}\]

    Next we will condense logarithmic expressions. As we will see, it is important to be able to combine an expression involving logarithms into a single logarithm with coefficient \(1\). This will be one of the first steps when solving logarithmic equations.

    Example \(\PageIndex{11}\):

    Write as a single logarithm with coefficient \(1\): \(3 \log _{3} x-\log _{3} y+2 \log _{3} 5\).

    Solution

    Begin by rewriting all of the logarithmic terms with coefficient 1. Use the power rule to do this. Then use the product and quotient rules to simplify further.

    \(\begin{aligned} 3 \log _{3} x-\log _{3} y+2 \log _{3} 5 &=\left\{\log _{3} x^{3}-\log _{3} y\right\}+\log _{3} 5^{2}\quad\color{Cerulean}{quotient\:property} \\ &=\left\{\log _{3}\left(\frac{x^{3}}{y}\right)+\log _{3} 25\right\} \quad\quad\:\color{Cerulean}{product\:property}\\ &=\log _{3}\left(\frac{x^{3}}{y} \cdot 25\right) \\ &=\log _{3}\left(\frac{25 x^{3}}{y}\right) \end{aligned}\)

    Answer

    \[\log _{3}\left(\frac{25 x^{3}}{y}\right) \nonumber\]

    Example \(\PageIndex{12}\):

    Write as a single logarithm with coefficient \(1\): \(\frac{1}{2} \ln x-3 \ln y-\ln z\).

    Solution

    Begin by writing the coefficients of the logarithms as powers of their argument, after which we will apply the quotient rule twice working from left to right.

    \(\begin{aligned} \frac{1}{2} \ln x-3 \ln y-\ln z &=\ln x^{1 / 2}-\ln y^{3}-\ln z \\ &=\ln \left(\frac{x^{1 / 2}}{y^{3}}\right)-\ln z \\ &=\ln \left(\frac{x^{1 / 2}}{y^{3}} \div z\right) \\ &=\ln \left(\frac{x^{1 / 2}}{y^{3}} \cdot \frac{1}{z}\right) \\ &=\ln \left(\frac{x^{1 / 2}}{y^{3} z}\right) \quad \text { or } \quad=\ln \left(\frac{\sqrt{x}}{y^{3} z}\right) \end{aligned}\)

    Answer

    \[\ln \left(\frac{\sqrt{x}}{y^{3}z}\right) \nonumber\]

    Exercise \(\PageIndex{3}\)

    Write as a single logarithm with coefficient \(1\): \(3 \log (x+y)-6 \log z+2 \log 5\)

    Answer

    \(\log \left(\frac{25(x+y)^{3}}{z^{6}}\right)\)

    http://www.youtube.com/v/o1AVbmS9AU8

    Key Takeaways

    • Given any base \(b > 0\) and \(b ≠ 1\), we can say that \(log_{b} 1 = 0\), \(log_{b} b = 1\), \(log_{1/b} b = −1\) and that \(log_{b} (\frac{1}{b}) = −1\).
    • The inverse properties of the logarithm are \(log_{b} b^{x} = x\) and \(b^{log_{b} x} = x\) where \(x > 0\).
    • The product property of the logarithm allows us to write a product as a sum: \(\log _{b}(x y)=\log _{b} x+\log _{b} y\).
    • The quotient property of the logarithm allows us to write a quotient as a difference: \(\log _{b}\left(\frac{x}{y}\right)=\log _{b} x-\log _{b} y\).
    • The power property of the logarithm allows us to write exponents as coefficients: \(\log _{b} x^{n}=n \log _{b} x\).
    • Since the natural logarithm is a base-\(e\) logarithm, \(\ln x=\log _{e} x\), all of the properties of the logarithm apply to it.
    • We can use the properties of the logarithm to expand logarithmic expressions using sums, differences, and coefficients. A logarithmic expression is completely expanded when the properties of the logarithm can no further be applied.
    • We can use the properties of the logarithm to combine expressions involving logarithms into a single logarithm with coefficient \(1\). This is an essential skill to be learned in this chapter.

    Exercise \(\PageIndex{4}\)

    Evaluate:

    1. \(\log _{7} 1\)
    2. \(\log _{1 / 2} 2\)
    3. \(\log 10^{14}\)
    4. \(\log 10^{-23}\)
    5. \(\log _{3} 3^{10}\)
    6. \(\log _{6} 6\)
    7. \(\ln e^{7}\)
    8. \(\ln \left(\frac{1}{e}\right)\)
    9. \(\log _{1 / 2}\left(\frac{1}{2}\right)\)
    10. \(\log _{1 / 5} 5\)
    11. \(\log _{3 / 4}\left(\frac{4}{3}\right)\)
    12. \(\log _{2 / 3} 1\)
    13. \(2^{\log _{2} 100}\)
    14. \(3^{\log _{3} 1}\)
    15. \(10^{\log 18}\)
    16. \(e^{\ln 23}\)
    17. \(e^{\ln x^{2}}\)
    18. \(e^{\ln e^{x}}\)
    Answer

    1. \(0\)

    3. \(14\)

    5. \(10\)

    7. \(7\)

    9. \(1\)

    11. \(−1\)

    13. \(100\)

    15. \(18\)

    17. \(x^{2}\)

    Exercise \(\PageIndex{5}\)

    Find \(a\):

    1. \(\ln a=1\)
    2. \(\log a=-1\)
    3. \(\log _{9} a=-1\)
    4. \(\log _{12} a=1\)
    5. \(\log _{2} a=5\)
    6. \(\log a=13\)
    7. \(2^{a}=7\)
    8. \(e^{a}=23\)
    9. \(\log _{a} 4^{5}=5\)
    10. \(\log _{a} 10=1\)
    Answer

    1. \(e\)

    3. \(\frac{1}{9}\)

    5. \(2^{5}=32\)

    7. \(\log _{2} 7\)

    9. \(4\)

    Exercise \(\PageIndex{6}\)

    Expand completely.

    1. \(\log _{4}(x y)\)
    2. \(\log (6 x)\)
    3. \(\log _{3}\left(9 x^{2}\right)\)
    4. \(\log _{2}\left(32 x^{7}\right)\)
    5. \(\ln \left(3 y^{2}\right)\)
    6. \(\log \left(100 x^{2}\right)\)
    7. \(\log _{2}\left(\frac{x}{y^{2}}\right)\)
    8. \(\log _{5}\left(\frac{25}{x}\right)\)
    9. \(\log \left(10 x^{2} y^{3}\right)\)
    10. \(\log _{2}\left(2 x^{4} y^{5}\right)\)
    11. \(\log _{3}\left(\frac{x^{3}}{y z^{2}}\right)\)
    12. \(\log \left(\frac{x}{y^{3} z^{2}}\right)\)
    13. \(\log _{5}\left(\frac{1}{x^{2} y z}\right)\)
    14. \(\log _{4}\left(\frac{1}{16 x^{2} z^{3}}\right)\)
    15. \(\log _{6}\left[36(x+y)^{4}\right]\)
    16. \(\ln \left[e^{4}(x-y)^{3}\right]\)
    17. \(\log _{7}(2 \sqrt{x y})\)
    18. \(\ln (2 x \sqrt{y})\)
    19. \(\log _{3}\left(\frac{x^{2} \sqrt[3]{y}}{z}\right)\)
    20. \(\log \left(\frac{2(x+y)^{3}}{z^{2}}\right)\)
    21. \(\log \left(\frac{100 x^{3}}{(y+10)^{3}}\right)\)
    22. \(\log _{7}\left(\frac{x}{\sqrt[5]{(y+z)^{3}}}\right)\)
    23. \(\log _{5}\left(\frac{x^{3}}{\sqrt[3]{y z^{2}}}\right)\)
    24. \(\log \left(\frac{x^{2}}{\sqrt[5]{y^{3} z^{2}}}\right)\)
    Answer

    1. \(\log _{4} x+\log _{4} y\)

    3. \(2+2 \log _{3} x\)

    5. \(\ln 3+2 \ln y\)

    7. \(\log _{2} x-2 \log _{2} y\)

    9. \(1+2 \log x+3 \log y\)

    11. \(3 \log _{3} x-\log _{3} y-2 \log _{3} z\)

    13. \(-2 \log _{5} x-\log _{5} y-\log _{5} z\)

    15. \(2+4 \log _{6}(x+y)\)

    17. \(\log _{7} 2+\frac{1}{2} \log _{7} x+\frac{1}{2} \log _{7} y\)

    19. \(2 \log _{3} x+\frac{1}{3} \log _{3} y-\log _{3} z\)

    21. \(2+3 \log x-3 \log (y+10)\)

    23. \(3 \log _{5} x-\frac{1}{3} \log _{5} y-\frac{2}{3} \log _{5} z\)

    Exercise \(\PageIndex{7}\)

    Given \(\log _{3} x=a, \log _{3} y=b\), and \(\log _{3} z=c\), write the following logarithms in terms of \(a, b\), and and \(c\).

    1. \(\log _{3}\left(27 x^{2} y^{3} z\right)\)
    2. \(\log _{3}\left(x y^{3} \sqrt{z}\right)\)
    3. \(\log _{3}\left(\frac{9 x^{2} y}{z^{3}}\right)\)
    4. \(\log _{3}\left(\frac{\sqrt[3]{x}}{y z^{2}}\right)\)
    Answer

    1. \(3+2 a+3 b+c\)

    3. \(2+2 a+b-3 c\)

    Exercise \(\PageIndex{8}\)

    Given \(\log _{b} 2=0.43, \log _{b} 3=0.68\), and \(\log _{b} 7=1.21\), calculate the following. (Hint: Expand using sums, differences, and quotients of the factors \(2, 3\), and \(7\).)

    1. \(\log _{b} 42\)
    2. \(\log _{b}(36)\)
    3. \(\log _{b}\left(\frac{28}{9}\right)\)
    4. \(\log _{b} \sqrt{21}\)
    Answer

    1. \(2.32\)

    3. \(0.71\)

    Exercise \(\PageIndex{9}\)

    Expand using the properties of the logarithm and then approximate using a calculator to the nearest tenth.

    1. \(\log \left(3.10 \times 10^{25}\right)\)
    2. \(\log \left(1.40 \times 10^{-33}\right)\)
    3. \(\ln \left(6.2 e^{-15}\right)\)
    4. \(\ln \left(1.4 e^{22}\right)\)
    Answer

    1. \(\log (3.1)+25 \approx 25.5\)

    3. \(\ln (6.2)-15 \approx-13.2\)

    Exercise \(\PageIndex{10}\)

    Write as a single logarithm with coefficient \(1\).

    1. \(\log x+\log y\)
    2. \(\log _{3} x-\log _{3} y\)
    3. \(\log _{2} 5+2 \log _{2} x+\log _{2} y\)
    4. \(\log _{3} 4+3 \log _{3} x+\frac{1}{2} \log _{3} y\)
    5. \(3 \log _{2} x-2 \log _{2} y+\frac{1}{2} \log _{2} z\)
    6. \(4 \log x-\log y-\log 2\)
    7. \(\log 5+3 \log (x+y)\)
    8. \(4 \log _{5}(x+5)+\log _{5} y\)
    9. \(\ln x-6 \ln y+\ln z\)
    10. \(\log _{3} x-2 \log _{3} y+5 \log _{3} z\)
    11. \(7 \log x-\log y-2 \log z\)
    12. \(2 \ln x-3 \ln y-\ln z\)
    13. \(\frac{2}{3} \log _{3} x-\frac{1}{2}\left(\log _{3} y+\log _{3} z\right)\)
    14. \(\frac{1}{5}\left(\log _{7} x+2 \log _{7} y\right)-2 \log _{7}(z+1)\)
    15. \(1+\log _{2} x-\frac{1}{2} \log _{2} y\)
    16. \(2-3 \log _{3} x+\frac{1}{3} \log _{3} y\)
    17. \(\frac{1}{3} \log _{2} x+\frac{2}{3} \log _{2} y\)
    18. \(-2 \log _{5} x+\frac{3}{5} \log _{5} y\)
    19. \(-\ln 2+2 \ln (x+y)-\ln z\)
    20. \(-3 \ln (x-y)-\ln z+\ln 5\)
    21. \(\frac{1}{3}(\ln x+2 \ln y)-(3 \ln 2+\ln z)\)
    22. \(4 \log 2+\frac{2}{3} \log x-4 \log (y+z)\)
    23. \(\log _{2} 3-2 \log _{2} x+\frac{1}{2} \log _{2} y-4 \log _{2} z\)
    24. \(2 \log _{5} 4-\log _{5} x-3 \log _{5} y+\frac{2}{3} \log _{5} z\)
    Answer

    1. \(\log (x y)\)

    3. \(\log _{2}\left(5 x^{2} y\right)\)

    5. \(\log _{2}\left(\frac{x^{3} \sqrt{z}}{y^{2}}\right)\)

    7. \(\log \left[5(x+y)^{3}\right]\)

    9. \(\ln \left(\frac{x z}{y^{6}}\right)\)

    11. \(\log \left(\frac{x^{7}}{y z^{2}}\right)\)

    13. \(\log _{3}\left(\frac{\sqrt[3]{x^{2}}}{\sqrt{y z}}\right)\)

    15. \(\log _{2}\left(\frac{2 x}{\sqrt{y}}\right)\)

    17. \(\log _{2}\left(\sqrt[3]{x y^{2}}\right)\)

    19. \(\ln \left(\frac{(x+y)^{2}}{2 z}\right)\)

    21. \(\ln \left(\frac{\sqrt[3]{x y^{2}}}{8 z}\right)\)

    23. \(\log _{2}\left(\frac{3 \sqrt{y}}{x^{2} z^{4}}\right)\)

    Exercise \(\PageIndex{11}\)

    Express as a single logarithm and simplify.

    1. \(\log (x+1)+\log (x-1)\)
    2. \(\log _{2}(x+2)+\log _{2}(x+1)\)
    3. \(\ln \left(x^{2}+2 x+1\right)-\ln (x+1)\)
    4. \(\ln \left(x^{2}-9\right)-\ln (x+3)\)
    5. \(\log _{5}\left(x^{3}-8\right)-\log _{5}(x-2)\)
    6. \(\log _{3}\left(x^{3}+1\right)-\log _{3}(x+1)\)
    7. \(\log x+\log (x+5)-\log \left(x^{2}-25\right)\)
    8. \(\log (2 x+1)+\log (x-3)-\log \left(2 x^{2}-5 x-3\right)\)
    Answer

    1. \(\log \left(x^{2}-1\right)\)

    3. \(\ln (x+1)\)

    5. \(\log _{5}\left(x^{2}+2 x+4\right)\)

    7. \(\log \left(\frac{x}{x-5}\right)\)

    Footnotes

    11Given \(b > 0\) we have \(\log _{b} b^{x}=x\) and \(b^{\log _{b} x}=x\) when \(x > 0\).

    12\(\log _{b}(x y)=\log _{b} x+\log _{b} y\); the logarithm of a product is equal to the sum of the logarithm of the factors.

    13\(\log _{b}\left(\frac{x}{y}\right)=\log _{b} x-\log _{b} y\); the logarithm of a quotient is equal to the difference of the logarithm of the numerator and the logarithm of the denominator.

    14\(\log _{b} x^{n}=n \log _{b} x\); the logarithm of a quantity raised to a power is equal to that power times the logarithm of the quantity.]