1. logb2+logb2+logb2+logbk=3logb2+logbk
2. log3(x+3)−log3(x−1)−log3(x−2)
3. 2lnx
4. −2ln(x)
5. log316
6. 2logx+3logy−4logz
7. 32lnx
8. 21ln(x−1)+ln(2x+1)−ln(x+3)−ln(x−3)
9. log(4⋅63⋅5); can also be written log(85) by reducing the fraction to lowest terms.
10. log((7x−1)5(x−1)3x)
11. log(2x+3)4x12(x+5)4; this answer could also be written log((2x+3)x3(x+5))4.
12. The pH increases by about 0.301.
13. ln0.5ln8
14. ln5ln100≈1.60944.6051=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(xn1)=n1logb(x).
3. logb(2)+logb(7)+logb(x)+logb(y)
5. logb(13)−logb(17)
7. −kln(4)
9. ln(7xy)
11. logb(4)
13. logb(7)
15. 15log(x)+13log(y)−19log(z)
17. 23log(x)−2log(y)
19. 38log(x)+314log(y)
21. ln(2x7)
23. log(yxz3)
25. log7(15)=ln(7)ln(15)
27. log11(5)=log5(11)log5(5)=b1
29. log11(116)=log5(11)log5(116)=log5(11)log5(6)−log5(11)=ba−b=ba−1
31. 3
33. 2.81359
35. 0.93913
37. –2.23266
39. x = 4; By the quotient rule: log6(x+2)−log6(x−3)=log6(x−3x+2)=1.
Rewriting as an exponential equation and solving for x:
Checking, we find that log6(4+2)−log6(4−3)=log6(6)−log6(1) is defined, so x = 4.
41. Let b and n be positive integers greater than 1. Then, by the change-of-base formula, logb(n)=logn(b)logn(n)=logn(b)1.
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Precalculus.Provided by: OpenStaxAuthored 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/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175..