Solutions to Try Its

1. $\left(2,\infty \right)$

2. $\left(5,\infty \right)$

3. The domain is $\left(0,\infty \right)$, the range is $\left(-\infty ,\infty \right)$, and the vertical asymptote is x = 0.

4. The domain is $\left(-4,\infty \right)$, the range $\left(-\infty ,\infty \right)$, and the asymptote x = –4.

5. The domain is $\left(0,\infty \right)$, the range is $\left(-\infty ,\infty \right)$, and the vertical asymptote is x = 0.

6. The domain is $\left(0,\infty \right)$, the range is $\left(-\infty ,\infty \right)$, and the vertical asymptote is x = 0.

7. The domain is $\left(2,\infty \right)$, the range is $\left(-\infty ,\infty \right)$, and the vertical asymptote is x = 2.

8. The domain is $\left(-\infty ,0\right)$, the range is $\left(-\infty ,\infty \right)$, and the vertical asymptote is x = 0.

9. $x\approx 3.049$

10. x = 1

11. $f\left(x\right)=2\mathrm{ln}\left(x+3\right)-1$

Solutions to Odd-Numbered Exercises

1. Since the functions are inverses, their graphs are mirror images about the line y = x. So for every point $\left(a,b\right)$ on the graph of a logarithmic function, there is a corresponding point $\left(b,a\right)$ on the graph of its inverse exponential function.

3. Shifting the function right or left and reflecting the function about the y-axis will affect its domain.

5. No. A horizontal asymptote would suggest a limit on the range, and the range of any logarithmic function in general form is all real numbers.

7. Domain: $\left(-\infty ,\frac{1}{2}\right)$; Range: $\left(-\infty ,\infty \right)$

9. Domain: $\left(-\frac{17}{4},\infty \right)$; Range: $\left(-\infty ,\infty \right)$

11. Domain: $\left(5,\infty \right)$; Vertical asymptote: x = 5

13. Domain: $\left(-\frac{1}{3},\infty \right)$; Vertical asymptote: $x=-\frac{1}{3}$

15. Domain: $\left(-3,\infty \right)$; Vertical asymptote: x = –3

17. Domain: $\left(\frac{3}{7},\infty \right)$; Vertical asymptote: $x=\frac{3}{7}$ ; End behavior: as $x\to {\left(\frac{3}{7}\right)}^{+},f\left(x\right)\to -\infty$ and as $x\to \infty ,f\left(x\right)\to \infty$

19. Domain: $\left(-3,\infty \right)$ ; Vertical asymptote: x = –3; End behavior: as $x\to -{3}^{+}$ , $f\left(x\right)\to -\infty$ and as $x\to \infty$ , $f\left(x\right)\to \infty$

21. Domain: $\left(1,\infty \right)$; Range: $\left(-\infty ,\infty \right)$; Vertical asymptote: x = 1; x-intercept: $\left(\frac{5}{4},0\right)$; y-intercept: DNE

23. Domain: $\left(-\infty ,0\right)$; Range: $\left(-\infty ,\infty \right)$; Vertical asymptote: x = 0; x-intercept: $\left(-{e}^{2},0\right)$; y-intercept: DNE

25. Domain: $\left(0,\infty \right)$; Range: $\left(-\infty ,\infty \right)$; Vertical asymptote: x = 0; x-intercept: $\left({e}^{3},0\right)$; y-intercept: DNE

27. B

29. C

31. B

33. C

35.

37.

39. C

41.

43.

45.

47. $f\left(x\right)={\mathrm{log}}_{2}\left(-\left(x - 1\right)\right)$

49. $f\left(x\right)=3{\mathrm{log}}_{4}\left(x+2\right)$

51. x = 2

53. $x\approx \text{2}\text{.303}$

55. $x\approx -0.472$

57. The graphs of $f\left(x\right)={\mathrm{log}}_{\frac{1}{2}}\left(x\right)$ and $g\left(x\right)=-{\mathrm{log}}_{2}\left(x\right)$ appear to be the same; Conjecture: for any positive base $b\ne 1$, ${\mathrm{log}}_{b}\left(x\right)=-{\mathrm{log}}_{\frac{1}{b}}\left(x\right)$.

59. Recall that the argument of a logarithmic function must be positive, so we determine where $\frac{x+2}{x - 4}>0$ . From the graph of the function $f\left(x\right)=\frac{x+2}{x - 4}$, note that the graph lies above the x-axis on the interval $\left(-\infty ,-2\right)$ and again to the right of the vertical asymptote, that is $\left(4,\infty \right)$. Therefore, the domain is $\left(-\infty ,-2\right)\cup \left(4,\infty \right)$.