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 ${b}^{y}$ can also be written as the logarithm, ${\mathrm{log}}_{b}x$, 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 ${b}^{y}=x$\\, 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 ${\mathrm{log}}_{e}\left(x\right)$, the notation used is $\mathrm{ln}\left(x\right)$.

7. ${a}^{c}=b$

9. ${x}^{y}=64$

11. ${15}^{b}=a$

13. ${13}^{a}=142$

15. ${e}^{n}=w$

17. ${\text{log}}_{c}\left(k\right)=d$

19. ${\mathrm{log}}_{19}y=x$

21. ${\mathrm{log}}_{n}\left(103\right)=4$

23. ${\mathrm{log}}_{y}\left(\frac{39}{100}\right)=x$

25. $\text{ln}\left(h\right)=k$

27. $x={2}^{-3}=\frac{1}{8}$

29. $x={3}^{3}=27$

31. $x={9}^{\frac{1}{2}}=3$

33. $x={6}^{-3}=\frac{1}{216}$

35. $x={e}^{2}$

37. 32

39. 1.06

41. 14.125

43. $\frac{1}{2}$

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

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

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

65. 2