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Method 2: 51-3 = 10 => log, (5-) = log, 10 t-3 = log, 1C by(2) a) 4·73t-1 = 7 35t b) 7 – 4 = 6 c) In(4 – 5x) = 3 d) 5log3(4x) – 8 = 3 e) 3logs(1-x) = 5 f) 3 In(13²+3*) – 5 = 2

Method 2: 51-3 = 10 => log, (5-) = log, 10 t-3 = log, 1C by(2) a) 4·73t-1 = 7 35t b) 7 – 4 = 6 c) In(4 – 5x) = 3 d) 5log3(4x) – 8 = 3 e) 3logs(1-x) = 5 f) 3 In(13²+3*) – 5 = 2

Trigonometry (11th Edition)
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Method 2: 51-3 = 10 => log, (5-3 )= log, 10 t- 3 = log, 10=t = %3D by(2) a) 4·73t-1 = 7 35t b) 7 – 4 = 6 - c) In(4 – 5x) = 3 - d) 5log3(4x) – 8 = 3 e) 3log5(1-x) = 5 %3D f) 3 ln(13²+3*) – 5 = 2
Transcribed Image Text:Method 2: 51-3 = 10 => log, (5-3 )= log, 10 t- 3 = log, 10=t = %3D by(2) a) 4·73t-1 = 7 35t b) 7 – 4 = 6 - c) In(4 – 5x) = 3 - d) 5log3(4x) – 8 = 3 e) 3log5(1-x) = 5 %3D f) 3 ln(13²+3*) – 5 = 2
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Rewrite each function in the form 

y = A0eat,

 for appropriate constants A0 and a. (Round A0 and a to four decimal places as necessary.)

y = −5(1.569)t − 5

 

 

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