We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

Study Guides > MATH 1314: College Algebra

Solutions

Solutions to Try Its

1. logb2+logb2+logb2+logbk=3logb2+logbk{\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. log3(x+3)log3(x1)log3(x2){\mathrm{log}}_{3}\left(x+3\right)-{\mathrm{log}}_{3}\left(x - 1\right)-{\mathrm{log}}_{3}\left(x - 2\right)\\ 3. 2lnx2\mathrm{ln}x\\ 4. 2ln(x)-2\mathrm{ln}\left(x\right)\\ 5. log316{\mathrm{log}}_{3}16\\ 6. 2logx+3logy4logz2\mathrm{log}x+3\mathrm{log}y - 4\mathrm{log}z\\ 7. 23lnx\frac{2}{3}\mathrm{ln}x\\ 8. 12ln(x1)+ln(2x+1)ln(x+3)ln(x3)\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. log(3546)\mathrm{log}\left(\frac{3\cdot 5}{4\cdot 6}\right)\\; can also be written log(58)\mathrm{log}\left(\frac{5}{8}\right)\\ by reducing the fraction to lowest terms. 10. log(5(x1)3x(7x1))\mathrm{log}\left(\frac{5{\left(x - 1\right)}^{3}\sqrt{x}}{\left(7x - 1\right)}\right)\\ 11. logx12(x+5)4(2x+3)4\mathrm{log}\frac{{x}^{12}{\left(x+5\right)}^{4}}{{\left(2x+3\right)}^{4}}\\; this answer could also be written log(x3(x+5)(2x+3))4\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. ln8ln0.5\frac{\mathrm{ln}8}{\mathrm{ln}0.5}\\ 14. ln100ln54.60511.6094=2.861\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, logb(x1n)=1nlogb(x){\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}\left(x\right)\\. 3. logb(2)+logb(7)+logb(x)+logb(y){\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. logb(13)logb(17){\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)\\ 7. kln(4)-k\mathrm{ln}\left(4\right)\\ 9. ln(7xy)\mathrm{ln}\left(7xy\right)\\ 11. logb(4){\mathrm{log}}_{b}\left(4\right)\\ 13. logb(7){\text{log}}_{b}\left(7\right)\\ 15. 15log(x)+13log(y)19log(z)15\mathrm{log}\left(x\right)+13\mathrm{log}\left(y\right)-19\mathrm{log}\left(z\right)\\ 17. 32log(x)2log(y)\frac{3}{2}\mathrm{log}\left(x\right)-2\mathrm{log}\left(y\right)\\ 19. 83log(x)+143log(y)\frac{8}{3}\mathrm{log}\left(x\right)+\frac{14}{3}\mathrm{log}\left(y\right)\\ 21. ln(2x7)\mathrm{ln}\left(2{x}^{7}\right)\\ 23. log(xz3y)\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)\\ 25. log7(15)=ln(15)ln(7){\mathrm{log}}_{7}\left(15\right)=\frac{\mathrm{ln}\left(15\right)}{\mathrm{ln}\left(7\right)}\\ 27. log11(5)=log5(5)log5(11)=1b{\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}\\ 29. log11(611)=log5(611)log5(11)=log5(6)log5(11)log5(11)=abb=ab1{\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: log6(x+2)log6(x3)=log6(x+2x3)=1{\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:

{61=x+2x30=x+2x360=x+2x36(x3)(x3)0=x+26x+18x30=x4x3 x=4\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 log6(4+2)log6(43)=log6(6)log6(1){\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, logb(n)=logn(n)logn(b)=1logn(b){\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)}\\.

Licenses & Attributions