Logarithmic Properties

1. [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]

2. [latex]{\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)[/latex]

3. [latex]-k\mathrm{ln}\left(4\right)[/latex]

4. [latex]\mathrm{ln}\left(7xy\right)[/latex]

5. [latex]{\mathrm{log}}_{b}\left(4\right)[/latex]

6. [latex]{\text{log}}_{b}\left(7\right)[/latex]

7. [latex]15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)[/latex]

8. [latex]\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)[/latex]

9. [latex]\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)[/latex]

10. [latex]\mathrm{ln}\left(2{x}^{7}\right)[/latex]

11. [latex]\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)[/latex]

12. [latex]{\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}[/latex]

13. [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]

14. 3

15. 2.81359

16. 0.93913

17. –2.23266

18. 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.

Exponential and Logarithmic Equations

1. An extraneous solution occurs when solving the equation results in a value that does not satisfy the original equation. It can be recognized by substituting the solution back into the original equation and seeing that it makes the equation false or violates a domain restriction (for example, giving a negative input to a logarithm or an even root, or causing division by zero).

2. The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

3. [latex]x=-\frac{1}{3}[/latex]

4. n = –1

5. [latex]b=\frac{6}{5}[/latex]

6. x = 10

7. No solution

8. [latex]p=\mathrm{log}\left(\frac{17}{8}\right)-7[/latex]

9. [latex]x=\mathrm{ln}12[/latex]

10. [latex]x=\mathrm{ln}\left(3\right)[/latex]

11. n = 49

12. [latex]k=\frac{1}{36}[/latex]

13. [latex]x=\frac{9-e}{8}[/latex]

14. n = 1

15. No solution

16. No solution

17. [latex]x=\pm \frac{10}{3}[/latex]

18. x = 9

19. [latex]x=\frac{{e}^{2}}{3}\approx 2.5[/latex]

20. x = –5
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