By Jon Rogawski
Instructor's ideas handbook (Part 2) for Thomas' Calculus 11e and Thomas' Calculus: Early Transcendentals 11e
Read Online or Download Instructor's Solution Manuals to Calculus: Early Transcendentals PDF
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Additional info for Instructor's Solution Manuals to Calculus: Early Transcendentals
Example text
X3 x2 + 1 = 2 or 2x 3 − x 2 − 1 = 0; x = 1 is the only real root to this equation. Substituting x = 1 into the x3 original equation, we find Therefore, ln 2 − 3 ln 1 = ln 2 − 0 = ln 2 as needed. Hence, x = 1 is the only solution. In Exercises 29–34, solve for the unknown. 29. 7e5t = 100 solution Divide the equation by 7 and then take the natural logarithm of both sides. This gives 5t = ln March 30, 2011 100 7 or t = 1 ln 5 100 . 6 57 30. 6e−4t = 2 solution Divide the equation by 6 and then take the natural logarithm of both sides.
Csc θ = 2 solution θ = π6 , 5π 6 √ 3 13. sin x = 2 solution x = π3 , 2π 3 14. sec t = 2 solution t = π3 , 5π 3 15. Fill in the following table of values: solution π 6 θ π 4 π 3 π 2 2π 3 3π 4 5π 6 tan θ sec θ solution π 6 π 4 tan θ 1 √ 3 sec θ 2 √ 3 θ 1 π 3 √ 3 π 2 √ 2 2 sin θ cos θ + + 3π 4 5π 6 und 2π 3 √ − 3 −1 1 −√ 3 und −2 √ − 2 2 −√ 3 16. Complete the following table of signs: θ 0<θ < π 2 π <θ <π 2 π <θ < 3π 2 3π < θ < 2π 2 March 30, 2011 tan θ cot θ sec θ csc θ 33 34 CHAPTER 1 PRECALCULUS REVIEW solution θ 0<θ < π 2 π <θ <π 2 π <θ < 3π 2 3π < θ < 2π 2 sin cos tan cot sec csc + + + + + + + − − − − + − − + + − − − + − − + − 17.
K 3/2 = 27 solution 10. b2 x+1 solution Raise both sides of the equation to the two-thirds power. This gives k = (27)2/3 = (271/3 )2 = 32 = 9. = b−6 Rewrite (b2 )x+1 as b2(x+1) . Then 2(x + 1) = −6, and x = −4. In Exercises 11–26, calculate without using a calculator. 11. log3 27 solution log3 27 = log3 33 = 3 log3 3 = 3. 1 12. log5 25 solution log5 1 = log5 5−2 = −2 log5 5 = −2. 25 13. ln 1 solution ln 1 = 0. 14. log5 (54 ) solution log5 (54 ) = 4 log5 5 = 4. 15. log2 (25/3 ) solution log2 25/3 = 5 5 log2 2 = .