Solve equation/inequality for x where logeg * + log6g (36x + (3)) = 1 Note: Write your answer as a fraction (For example, write 1/2 but not 0.5)

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Tell me the right answer asap q5 read the note too please
Solve equation/inequality for x where log6g *+ log68 (36x + (3)) = 1
Note: Write your answer as a fraction (For example, write 1/2 but not 0.5)
Transcribed Image Text:Solve equation/inequality for x where log6g *+ log68 (36x + (3)) = 1 Note: Write your answer as a fraction (For example, write 1/2 but not 0.5)
Expert Solution
Step 1

log68x + log68(36x + 3) =1

log68x(36x + 3) =1     (here we use formula logx + logy = logxy)

x(36x + 3) = 68 (here antilog 681 =68)

36x2 + 3x - 68 = 0

using the Quadratic Formula where
a = 36, b = 3, and c = -68

x=b±b2 -4ac/2a
x=3±99/72
x=96/72        x=102/72
which becomes
x=1.33333
x=1.41667
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