## Solutions to Try Its

1. ${\mathrm{log}}_{b}2+{\mathrm{log}}_{b}2+{\mathrm{log}}_{b}2+{\mathrm{log}}_{b}k=3{\mathrm{log}}_{b}2+{\mathrm{log}}_{b}k$

2. ${\mathrm{log}}_{3}\left(x+3\right)-{\mathrm{log}}_{3}\left(x - 1\right)-{\mathrm{log}}_{3}\left(x - 2\right)$

3. $2\mathrm{ln}x$

4. $-2\mathrm{ln}\left(x\right)$

5. ${\mathrm{log}}_{3}16$

6. $2\mathrm{log}x+3\mathrm{log}y - 4\mathrm{log}z$

7. $\frac{2}{3}\mathrm{ln}x$

8. $\frac{1}{2}\mathrm{ln}\left(x - 1\right)+\mathrm{ln}\left(2x+1\right)-\mathrm{ln}\left(x+3\right)-\mathrm{ln}\left(x - 3\right)$

9. $\mathrm{log}\left(\frac{3\cdot 5}{4\cdot 6}\right)$; can also be written $\mathrm{log}\left(\frac{5}{8}\right)$ by reducing the fraction to lowest terms.

10. $\mathrm{log}\left(\frac{5{\left(x - 1\right)}^{3}\sqrt{x}}{\left(7x - 1\right)}\right)$

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

12. The pH increases by about 0.301.

13. $\frac{\mathrm{ln}8}{\mathrm{ln}0.5}$

14. $\frac{\mathrm{ln}100}{\mathrm{ln}5}\approx \frac{4.6051}{1.6094}=2.861$

## Solutions to Odd-Numbered Exercises

1. Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, ${\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}\left(x\right)$.

3. ${\mathrm{log}}_{b}\left(2\right)+{\mathrm{log}}_{b}\left(7\right)+{\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$

5. ${\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)$

7. $-k\mathrm{ln}\left(4\right)$

9. $\mathrm{ln}\left(7xy\right)$

11. ${\mathrm{log}}_{b}\left(4\right)$

13. ${\text{log}}_{b}\left(7\right)$

15. $15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)$

17. $\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)$

19. $\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)$

21. $\mathrm{ln}\left(2{x}^{7}\right)$

23. $\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)$

25. ${\mathrm{log}}_{7}\left(15\right)=\frac{\mathrm{ln}\left(15\right)}{\mathrm{ln}\left(7\right)}$

27. ${\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}$

29. ${\mathrm{log}}_{11}\left(\frac{6}{11}\right)=\frac{{\mathrm{log}}_{5}\left(\frac{6}{11}\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{{\mathrm{log}}_{5}\left(6\right)-{\mathrm{log}}_{5}\left(11\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{a-b}{b}=\frac{a}{b}-1$

31. 3

33. 2.81359

35. 0.93913

37. –2.23266

39. = 4; By the quotient rule: ${\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x - 3\right)={\mathrm{log}}_{6}\left(\frac{x+2}{x - 3}\right)=1$.

Rewriting as an exponential equation and solving for x:

$\begin{cases}{6}^{1}\hfill & =\frac{x+2}{x - 3}\hfill \\ 0\hfill & =\frac{x+2}{x - 3}-6\hfill \\ 0\hfill & =\frac{x+2}{x - 3}-\frac{6\left(x - 3\right)}{\left(x - 3\right)}\hfill \\ 0\hfill & =\frac{x+2 - 6x+18}{x - 3}\hfill \\ 0\hfill & =\frac{x - 4}{x - 3}\hfill \\ \text{ }x\hfill & =4\hfill \end{cases}$

Checking, we find that ${\mathrm{log}}_{6}\left(4+2\right)-{\mathrm{log}}_{6}\left(4 - 3\right)={\mathrm{log}}_{6}\left(6\right)-{\mathrm{log}}_{6}\left(1\right)$ is defined, so = 4.

41. Let b and n be positive integers greater than 1. Then, by the change-of-base formula, ${\mathrm{log}}_{b}\left(n\right)=\frac{{\mathrm{log}}_{n}\left(n\right)}{{\mathrm{log}}_{n}\left(b\right)}=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}$.