Logarithmic Properties

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. 1. ⓐ Simplify \(3∙{2}^{x}∙{2}^{3x}\)
   2. ⓑ Simplify \(\frac{{b}^{2}{b}^{6}b}{{b}^{4}{b}^{7}}\)
   3. ⓒ Simplify \({(a{b}^{2})}^{3}{a}^{5}{b}^{-6}\)
   Bonisa impendulo

   ⓐ

   Use the product property.	\(3∙{2}^{x+3x}\)
   Simplify.	\(3∙{2}^{4x}\)

   ⓑ

   Use the product property and multiply exponents.	\(\frac{{b}^{9}}{{b}^{11}}\)
   Use the quotient property and add exponents.	\({b}^{-2}=\frac{1}{{b}^{2}}\)

   ⓒ

   Use the power property and multiply exponents.	\({a}^{3}{b}^{6}{a}^{5}{b}^{-6}\)
   Use the product property and add exponents.	\({a}^{8}{b}^{0}\)
   Any base to the power of zero equals 1.	\({a}^{8}(1)={a}^{8}\)

2. \(3{b}^{5}∙2{b}^{12}\)

3. \(x∙{x}^{5}∙{x}^{7}\)

4. \(\frac{{b}^{15}{c}^{4}}{{b}^{4}c}\)

5. \(4{x}^{0}\)

6. \(12{x}^{-6}\)

7. \({(2{a}^{3})}^{3}{({a}^{4})}^{2}\)

8. \(\frac{{a}^{-3}{b}^{5}}{2{a}^{-6}}\)

9. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{4}(2{x}^{3}{y}^{2})\) . Simplify, if possible.

   Bonisa impendulo

   	\(\ {\text{log}}_{4}(2{x}^{3}{y}^{2})\)
   Use the Product Property, \({\text{log}}_{a}M\cdot N={\text{log}}_{a}M+{\text{log}}_{a}N\) .	\(\ {\text{log}}_{4}2+{\text{log}}_{4}{x}^{3}+{\text{log}}_{4}{y}^{2}\)
   Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\) , on the last two terms.	\(\ {\text{log}}_{4}2+3{\text{log}}_{4}x+2{\text{log}}_{4}y\)
   Simplify.	\(\ \frac{1}{2}+3{\text{log}}_{4}x+2{\text{log}}_{4}y\)
   	\(\ {\text{log}}_{4}(2{x}^{3}{y}^{2})=\frac{1}{2}+3{\text{log}}_{4}x+2{\text{log}}_{4}y\)

10. Use the Properties of Logarithms to expand the logarithm \({\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}\) . Simplify, if possible.

    Bonisa impendulo

    	\(\ {\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}\)
    Rewrite the radical with a rational exponent.	\(\ {\text{log}}_{2}{(\frac{{x}^{3}}{3{y}^{2}z})}^{\frac{1}{4}}\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\) .	\(\ \frac{1}{4}{\text{log}}_{2}(\frac{{x}^{3}}{3{y}^{2}z})\)
    Use the Quotient Property, \({\text{log}}_{a}M\cdot N={\text{log}}_{a}M-{\text{log}}_{a}N\) .	\(\ \frac{1}{4}({\text{log}}_{2}({x}^{3})-{\text{log}}_{2}(3{y}^{2}z))\)
    Use the Product Property, \({\text{log}}_{a}M\cdot N={\text{log}}_{a}M+{\text{log}}_{a}N\) , in the second term.	\(\ \frac{1}{4}({\text{log}}_{2}({x}^{3})-({\text{log}}_{2}3+{\text{log}}_{2}{y}^{2}+{\text{log}}_{2}z))\)
    Use the Power Property, \({\text{log}}_{a}{M}^{p}=p\ {\text{log}}_{a}M\) , inside the parentheses.	\(\ \frac{1}{4}(3{\text{log}}_{2}x-({\text{log}}_{2}3+2{\text{log}}_{2}y+{\text{log}}_{2}z))\)
    Simplify by distributing.	\(\ \frac{1}{4}(3{\text{log}}_{2}x-{\text{log}}_{2}3-2{\text{log}}_{2}y-{\text{log}}_{2}z)\)
    	\({\text{log}}_{2}\sqrt[4]{\frac{{x}^{3}}{3{y}^{2}z}}=\frac{1}{4}(3{\text{log}}_{2}x-{\text{log}}_{2}3-2{\text{log}}_{2}y-{\text{log}}_{2}z)\)

11. Use the Properties of Logarithms to condense the logarithm \({\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y\) . Simplify, if possible.

    Bonisa impendulo

    The log expressions all have the same base, 4.	\(\ {\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y\)
    The first two terms are added, so we use the Product Property, \({\text{log}}_{a}M+{\text{log}}_{a}N={\text{log}}_{a}M\cdot N\) .	\(\ {\text{log}}_{4}3x-{\text{log}}_{4}y\)
    Since the logs are subtracted, we use the Quotient Property, \({\text{log}}_{a}M-{\text{log}}_{a}N={\text{log}}_{a}\frac{M}{N}\) .	\(\ {\text{log}}_{4}\frac{3x}{y}\)
    	\(\ {\text{log}}_{4}3+{\text{log}}_{4}x-{\text{log}}_{4}y={\text{log}}_{4}\frac{3x}{y}\)

12. Use the properties of logarithms to expand: \({\log }_{3}(9{x}^{5}{y}^{4})\)

13. Use the properties of logarithms to expand: \({\log }_{5}\sqrt[3]{\frac{{x}^{5}}{25{y}^{3}z}}\)

14. Use the Properties of Logarithms to condense the logarithm: \({\log }_{b}5+{\log }_{b}c-{\log }_{b}b\)

15. Use the Properties of Logarithms to condense the logarithm: \(2{\log }_{3}x+3{\log }_{3}(x+1)\)

16. Expand \({\log }_{3}(30x(3x+4)).\)

    Bonisa impendulo

    We begin by factoring the argument completely, expressing \(30\) as a product of primes.

    \[{\log }_{3}(30x(3x+4))={\log }_{3}(2⋅3⋅5⋅x⋅(3x+4))\]

    Next we write the equivalent equation by summing the logarithms of each factor.

    \[{\log }_{3}(30x(3x+4))={\log }_{3}(2)+{\log }_{3}(3)+{\log }_{3}(5)+{\log }_{3}(x)+{\log }_{3}(3x+4)\]

17. Expand \({\log }_{b}(8k).\)

    Bonisa impendulo

    \({\log }_{b}2+{\log }_{b}2+{\log }_{b}2+{\log }_{b}k=3{\log }_{b}2+{\log }_{b}k\)

18. Expand \({\log }_{2}(\frac{15x(x-1)}{(3x+4)(2-x)}).\)

    Bonisa impendulo

    First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

    \[{\log }_{2}(\frac{15x(x-1)}{(3x+4)(2-x)})={\log }_{2}(15x(x-1))-{\log }_{2}((3x+4)(2-x))\]

    Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.

    \[\begin{array}{l}{\log }_{2}(15x(x-1))-{\log }_{2}((3x+4)(2-x))=[{\log }_{2}(3)+{\log }_{2}(5)+{\log }_{2}(x)+{\log }_{2}(x-1)]-[{\log }_{2}(3x+4)+{\log }_{2}(2-x)] \\ \ ={\log }_{2}(3)+{\log }_{2}(5)+{\log }_{2}(x)+{\log }_{2}(x-1)-{\log }_{2}(3x+4)-{\log }_{2}(2-x)\end{array}\]

19. Expand \({\log }_{3}(\frac{7{x}^{2}+21x}{7x(x-1)(x-2)}).\)

    Bonisa impendulo

    \({\log }_{3}(x+3)-{\log }_{3}(x-1)-{\log }_{3}(x-2)\)

20. Expand \({\log }_{2}{x}^{5}.\)

    Bonisa impendulo

    The argument is already written as a power, so we identify the exponent, 5, and the base, \(x,\) and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

    \[{\log }_{2}({x}^{5})=5{\log }_{2}x\]

21. Expand \(\ln {x}^{2}.\)

    Bonisa impendulo

    \(2\ln x\)

22. Expand \({\log }_{3}(25)\) using the power rule for logs.

    Bonisa impendulo

    Expressing the argument as a power, we get \({\log }_{3}(25)={\log }_{3}({5}^{2}).\)

    Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

    \[{\log }_{3}({5}^{2})=2{\log }_{3}(5)\]

23. Expand \(\ln (\frac{1}{{x}^{2}}).\)

    Bonisa impendulo

    \(-2\ln (x)\)

24. Rewrite \(4\ln (x)\) using the power rule for logs to a single logarithm with a leading coefficient of 1.

    Bonisa impendulo

    Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression \(4\ln (x),\) we identify the factor, 4, as the exponent and the argument, \(x,\) as the base, and rewrite the product as a logarithm of a power: \(4\ln (x)=\ln ({x}^{4}).\)

25. Rewrite \(2{\log }_{3}4\) using the power rule for logs to a single logarithm with a leading coefficient of 1.

    Bonisa impendulo

    \({\log }_{3}16\)

26. Rewrite \(\ln (\frac{{x}^{4}y}{7})\) as a sum or difference of logs.

    Bonisa impendulo

    First, because we have a quotient of two expressions, we can use the quotient rule:

    \[\ln (\frac{{x}^{4}y}{7})=\ln ({x}^{4}y)-\ln (7)\]

    Then seeing the product in the first term, we use the product rule:

    \[\ln ({x}^{4}y)-\ln (7)=\ln ({x}^{4})+\ln (y)-\ln (7)\]

    Finally, we use the power rule on the first term:

    \[\ln ({x}^{4})+\ln (y)-\ln (7)=4\ln (x)+\ln (y)-\ln (7)\]

27. Expand \(\log (\frac{{x}^{2}{y}^{3}}{{z}^{4}}).\)

    Bonisa impendulo

    \(2\log x+3\log y-4\log z\)

28. Expand \(\log (\sqrt{x}).\)

    Bonisa impendulo

    \[\begin{array}{ll}\log (\sqrt{x}) & =\log {x}^{(\frac{1}{2})} \\ & =\frac{1}{2}\log x\end{array}\]

29. Expand \(\ln (\sqrt[3]{{x}^{2}}).\)

    Bonisa impendulo

    \(\frac{2}{3}\ln x\)

30. Expand \({\log }_{6}(\frac{64{x}^{3}(4x+1)}{(2x-1)}).\)

    Bonisa impendulo

    We can expand by applying the Product and Quotient Rules.

    \[\begin{array}{lll}{\log }_{6}(\frac{64{x}^{3}(4x+1)}{(2x-1)}) & ={\log }_{6}64+{\log }_{6}{x}^{3}+{\log }_{6}(4x+1)-{\log }_{6}(2x-1) & \text{Apply the Quotient Rule}. \\ & ={\log }_{6}{2}^{6}+{\log }_{6}{x}^{3}+{\log }_{6}(4x+1)-{\log }_{6}(2x-1) & {\text{Simplify by writing 64 as 2}}^{6}. \\ & =6{\log }_{6}2+3{\log }_{6}x+{\log }_{6}(4x+1)-{\log }_{6}(2x-1) & \text{Apply the Power Rule}.\end{array}\]

31. Expand \(\ln (\frac{\sqrt{(x-1){(2x+1)}^{2}}}{({x}^{2}-9)}).\)

    Bonisa impendulo

    \(\frac{1}{2}\ln (x-1)+\ln (2x+1)-\ln (x+3)-\ln (x-3)\)

32. Write \({\log }_{3}(5)+{\log }_{3}(8)-{\log }_{3}(2)\) as a single logarithm.

    Bonisa impendulo

    Using the product and quotient rules

    \[{\log }_{3}(5)+{\log }_{3}(8)={\log }_{3}(5⋅8)={\log }_{3}(40)\]

    This reduces our original expression to

    \[{\log }_{3}(40)-{\log }_{3}(2)\]

    Then, using the quotient rule

    \[{\log }_{3}(40)-{\log }_{3}(2)={\log }_{3}(\frac{40}{2})={\log }_{3}(20)\]

33. Condense \(\log 3-\log 4+\log 5-\log 6.\)

    Bonisa impendulo

    \(\log (\frac{3⋅5}{4⋅6});\) can also be written \(\log (\frac{5}{8})\) by reducing the fraction to lowest terms.

34. Condense \({\log }_{2}({x}^{2})+\frac{1}{2}{\log }_{2}(x-1)-3{\log }_{2}({(x+3)}^{2}).\)

    Bonisa impendulo

    We apply the power rule first:

    \[{\log }_{2}({x}^{2})+\frac{1}{2}{\log }_{2}(x-1)-3{\log }_{2}({(x+3)}^{2})={\log }_{2}({x}^{2})+{\log }_{2}(\sqrt{x-1})-{\log }_{2}({(x+3)}^{6})\]

    Next we apply the product rule to the sum:

    \[{\log }_{2}({x}^{2})+{\log }_{2}(\sqrt{x-1})-{\log }_{2}({(x+3)}^{6})={\log }_{2}({x}^{2}\sqrt{x-1})-{\log }_{2}({(x+3)}^{6})\]

    Finally, we apply the quotient rule to the difference:

    \[{\log }_{2}({x}^{2}\sqrt{x-1})-{\log }_{2}({(x+3)}^{6})={\log }_{2}\frac{{x}^{2}\sqrt{x-1}}{{(x+3)}^{6}}\]

35. Rewrite \(\log (5)+0.5\log (x)-\log (7x-1)+3\log (x-1)\) as a single logarithm.

    Bonisa impendulo

    \(\log (\frac{5{(x-1)}^{3}\sqrt{x}}{(7x-1)})\)

36. Rewrite \(2\log x-4\log (x+5)+\frac{1}{x}\log (3x+5)\) as a single logarithm.

    Bonisa impendulo

    We apply the power rule first:

    \[2\log x-4\log (x+5)+\frac{1}{x}\log (3x+5)=\log ({x}^{2})-\log {(x+5)}^{4}+\log ({(3x+5)}^{{x}^{-1}})\]

    Next we rearrange and apply the product rule to the sum:

    \[\log ({x}^{2})-\log {(x+5)}^{4}+\log ({(3x+5)}^{{x}^{-1}})\]\[=\log ({x}^{2})+\log ({(3x+5)}^{{x}^{-1}})-\log {(x+5)}^{4}\]\[=\log ({x}^{2}{(3x+5)}^{{x}^{-1}})-\log {(x+5)}^{4}\]

    Finally, we apply the quotient rule to the difference:

    \[=\log ({x}^{2}{(3x+5)}^{{x}^{-1}})-{\log (x+5)}^{4}=\log \frac{{x}^{2}{(3x+5)}^{{x}^{-1}}}{{(x+5)}^{4}}\]

37. Condense \(4(3\log (x)+\log (x+5)-\log (2x+3)).\)

    Bonisa impendulo

    \(\log \frac{{x}^{12}{(x+5)}^{4}}{{(2x+3)}^{4}};\) this answer could also be written \(\log {(\frac{{x}^{3}(x+5)}{(2x+3)})}^{4}.\)

38. Recall that, in chemistry, \(\text{pH}=-\log [{H}^{+}].\) If the concentration of hydrogen ions in a liquid is doubled, what is the effect on pH?

    Bonisa impendulo

    Suppose \(C\) is the original concentration of hydrogen ions, and \(P\) is the original pH of the liquid. Then \(P=-\log (C).\) If the concentration is doubled, the new concentration is \(2C.\) Then the pH of the new liquid is

    \[\text{pH}=-\log (2C)\]

    Using the product rule of logs

    \[\text{pH}=-\log (2C)=-(\log (2)+\log (C))=-\log (2)-\log (C)\]

    Since \(P=-\log (C),\) the new pH is

    \[\text{pH}=P-\log (2)\approx P-0.301\]

    When the concentration of hydrogen ions is doubled, the pH decreases by about 0.301.

39. How does the pH change when the concentration of positive hydrogen ions is decreased by half?

    Bonisa impendulo

    The pH increases by about 0.301.

40. Change \({\log }_{5}3\) to a quotient of natural logarithms.

    Bonisa impendulo

    Because we will be expressing \({\log }_{5}3\) as a quotient of natural logarithms, the new base, \(n=e.\)

    We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.

    \[\begin{array}{ll}{\log }_{b}M & =\frac{\ln M}{\ln b} \\ {\log }_{5}3 & =\frac{\ln 3}{\ln 5}\end{array}\]

Zama wena

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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