EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Question
Chapter 3.4, Problem 55E
To determine
To find : solution to the logarithmic equation algebraically.
Expert Solution & Answer
Answer to Problem 55E
There is NO _ solution to the equation lnx−ln(x+1)=2
Explanation of Solution
Given information: lnx−ln(x+1)=2
Concept Involved:
The word “solve” means the process of finding the value of t that makes the equation true. In order to solve we need to undo whatever is done to t .
Formula Used
- lnm−lnn=ln(mn)
- If logarithmic equation is lnA=m , then the equivalent exponential equation is A=em
Calculation:
Use the logarithmic rule lnm−lnn=ln(mn) in left side to rewrite the equation
lnx(x+1)=2
Rewrite the logarithmic equation to exponential equation
x(x+1)=e2
Multiply (x+1) on both sides of the equation
(x+1) ⋅x(x+1)=e2(x+1)
Simplify fraction in left side of the equation
x=e2(x+1)
Distribute e2 in right side of the equation
x=e2x+e
Subtract e2x on both sides of the equation in the process of getting x in one side of the equation
x−e2x=e2x+e−e2x
Simplify in right side of the equation
x−e2x=e
Factor x in left side of the equation
(1−e2)x=e
Divide 1−e2 on both sides of the equation
(1−e2)x(1−e2)=e(1−e2)
Simplify fraction in both sides of the equation
x≈−0.4254591
Check for extraneous solution by substituting the result in the original equation.
x≈−0.4254591 makes the original equation FALSE
Conclusion:
There is NO _ solution to the equation lnx−ln(x+1)=2
Chapter 3 Solutions
EBK PRECALCULUS W/LIMITS
Ch. 3.1 - Prob. 1ECh. 3.1 - Prob. 2ECh. 3.1 - Prob. 3ECh. 3.1 - Prob. 4ECh. 3.1 - Prob. 5ECh. 3.1 - Prob. 6ECh. 3.1 - Prob. 7ECh. 3.1 - Prob. 8ECh. 3.1 - Prob. 9ECh. 3.1 - Prob. 10E
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