Solutions
Solutions to Try Its
1. [latex]{\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[/latex] 2. [latex]{\mathrm{log}}_{3}\left(x+3\right)-{\mathrm{log}}_{3}\left(x - 1\right)-{\mathrm{log}}_{3}\left(x - 2\right)[/latex] 3. [latex]2\mathrm{ln}x[/latex] 4. [latex]-2\mathrm{ln}\left(x\right)[/latex] 5. [latex]{\mathrm{log}}_{3}16[/latex] 6. [latex]2\mathrm{log}x+3\mathrm{log}y - 4\mathrm{log}z[/latex] 7. [latex]\frac{2}{3}\mathrm{ln}x[/latex] 8. [latex]\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)[/latex] 9. [latex]\mathrm{log}\left(\frac{3\cdot 5}{4\cdot 6}\right)[/latex]; can also be written [latex]\mathrm{log}\left(\frac{5}{8}\right)[/latex] by reducing the fraction to lowest terms. 10. [latex]\mathrm{log}\left(\frac{5{\left(x - 1\right)}^{3}\sqrt{x}}{\left(7x - 1\right)}\right)[/latex] 11. [latex]\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}}[/latex]; this answer could also be written [latex]\mathrm{log}{\left(\frac{{x}^{3}\left(x+5\right)}{\left(2x+3\right)}\right)}^{4}[/latex]. 12. The pH increases by about 0.301. 13. [latex]\frac{\mathrm{ln}8}{\mathrm{ln}0.5}[/latex] 14. [latex]\frac{\mathrm{ln}100}{\mathrm{ln}5}\approx \frac{4.6051}{1.6094}=2.861[/latex]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, [latex]{\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}\left(x\right)[/latex]. 3. [latex]{\mathrm{log}}_{b}\left(2\right)+{\mathrm{log}}_{b}\left(7\right)+{\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)[/latex] 5. [latex]{\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)[/latex] 7. [latex]-k\mathrm{ln}\left(4\right)[/latex] 9. [latex]\mathrm{ln}\left(7xy\right)[/latex] 11. [latex]{\mathrm{log}}_{b}\left(4\right)[/latex] 13. [latex]{\text{log}}_{b}\left(7\right)[/latex] 15. [latex]15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)[/latex] 17. [latex]\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)[/latex] 19. [latex]\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)[/latex] 21. [latex]\mathrm{ln}\left(2{x}^{7}\right)[/latex] 23. [latex]\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)[/latex] 25. [latex]{\mathrm{log}}_{7}\left(15\right)=\frac{\mathrm{ln}\left(15\right)}{\mathrm{ln}\left(7\right)}[/latex] 27. [latex]{\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}[/latex] 29. [latex]{\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[/latex] 31. 3 33. 2.81359 35. 0.93913 37. –2.23266 39. x = 4; By the quotient rule: [latex]{\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[/latex].Rewriting as an exponential equation and solving for x:
[latex]\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}[/latex]
Checking, we find that [latex]{\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)[/latex] is defined, so x = 4.
41. Let b and n be positive integers greater than 1. Then, by the change-of-base formula, [latex]{\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)}[/latex].Licenses & Attributions
CC licensed content, Shared previously
- Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution. License terms: Download For Free at : http://cnx.org/contents/[email protected]..
