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

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 44E
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Help with letter e and f

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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Follow-up Question

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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