Solutions 23: Logarithmic Functions

Solutions to Try Its

1. a. [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex] is equivalent to [latex]{10}^{6}=1,000,000[/latex]
b. [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex] is equivalent to [latex]{5}^{2}=25[/latex]

2. a. [latex]{3}^{2}=9[/latex] is equivalent to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
b. [latex]{5}^{3}=125[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
c. [latex]{2}^{-1}=\frac{1}{2}[/latex] is equivalent to [latex]{\text{log}}_{2}\left(\frac{1}{2}\right)=-1[/latex]

3. [latex]{\mathrm{log}}_{121}\left(11\right)=\frac{1}{2}[/latex] (recalling that [latex]\sqrt{121}={\left(121\right)}^{\frac{1}{2}}=11[/latex] )

4. [latex]{\mathrm{log}}_{2}\left(\frac{1}{32}\right)=-5[/latex]

5. It is not possible to take the logarithm of a negative number in the set of real numbers.

6. It is not possible to take the logarithm of a negative number in the set of real numbers.

Solutions to Odd-Numbered Exercises

1. A logarithm is an exponent. Specifically, it is the exponent to which a base b is raised to produce a given value. In the expressions given, the base b has the same value. The exponent, y, in the expression [latex]{b}^{y}[/latex] can also be written as the logarithm, [latex]{\mathrm{log}}_{b}x[/latex], and the value of x is the result of raising b to the power of y.

3. Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation [latex]{b}^{y}=x[/latex]\\, and then properties of exponents can be applied to solve for x.

5. The natural logarithm is a special case of the logarithm with base b in that the natural log always has base e. Rather than notating the natural logarithm as [latex]{\mathrm{log}}_{e}\left(x\right)[/latex], the notation used is [latex]\mathrm{ln}\left(x\right)[/latex].

7. [latex]{a}^{c}=b[/latex]

9. [latex]{x}^{y}=64[/latex]

11. [latex]{15}^{b}=a[/latex]

13. [latex]{13}^{a}=142[/latex]

15. [latex]{e}^{n}=w[/latex]

17. [latex]{\text{log}}_{c}\left(k\right)=d[/latex]

19. [latex]{\mathrm{log}}_{19}y=x[/latex]

21. [latex]{\mathrm{log}}_{n}\left(103\right)=4[/latex]

23. [latex]{\mathrm{log}}_{y}\left(\frac{39}{100}\right)=x[/latex]

25. [latex]\text{ln}\left(h\right)=k[/latex]

27. [latex]x={2}^{-3}=\frac{1}{8}[/latex]

29. [latex]x={3}^{3}=27[/latex]

31. [latex]x={9}^{\frac{1}{2}}=3[/latex]

33. [latex]x={6}^{-3}=\frac{1}{216}[/latex]

35. [latex]x={e}^{2}[/latex]

37. 32

39. 1.06

41. 14.125

43. [latex]\frac{1}{2}[/latex]

45. 4

47. –3

49. –12

51. 0

53. 10

55. 2.708

57. 0.151

59. No, the function has no defined value for = 0. To verify, suppose = 0 is in the domain of the function [latex]f\left(x\right)=\mathrm{log}\left(x\right)[/latex]. Then there is some number n such that [latex]n=\mathrm{log}\left(0\right)[/latex]. Rewriting as an exponential equation gives: [latex]{10}^{n}=0[/latex], which is impossible since no such real number n exists. Therefore, = 0 is not the domain of the function [latex]f\left(x\right)=\mathrm{log}\left(x\right)[/latex].

61. Yes. Suppose there exists a real number x such that [latex]\mathrm{ln}x=2[/latex]. Rewriting as an exponential equation gives [latex]x={e}^{2}[/latex], which is a real number. To verify, let [latex]x={e}^{2}[/latex]. Then, by definition, [latex]\mathrm{ln}\left(x\right)=\mathrm{ln}\left({e}^{2}\right)=2[/latex].

63. No; [latex]\mathrm{ln}\left(1\right)=0[/latex], so [latex]\frac{\mathrm{ln}\left({e}^{1.725}\right)}{\mathrm{ln}\left(1\right)}[/latex] is undefined.

65. 2