EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Chapter 3.4, Problem 58E
To determine

To find : solution to the logarithmic equation algebraically.

Expert Solution & Answer
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Answer to Problem 58E

The solution to the equation ln(x+1)−ln(x−2)=lnx is        x≈3.303    _

Explanation of Solution

Given information: ln(x+1)−ln(x−2)=lnx

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

  1. lnm−lnn=ln(mn)
  2. If logarithmic equation is lnA=m , then the equivalent exponential equation is A=em

Calculation:

Use the logarithmic rule lnm+lnn=ln(m⋅n) in left side to rewrite the equation

  ln(x+1)(x−2)=lnx

Since we have natural logarithm on both sides we can drop it.

  x+1x−2=x

Multiply (x−2) on both sides of the equation

  (x+1)(x−2)⋅(x−2)=x⋅(x−2)

Simplify fraction in left side of the equation

  x+1=x(x−2)

Distribute in right side of the equation

  x+1=x2−2x

Subtract x and 1 on both sides of the equation

  x+1−x−1=x2−2x−x−1

Simplify on both sides of the equation

  0=x2−3x−1

Use the symmetric property of equality to rewrite the equation

  x2−3x−1=0

When the quadratic equation of the form ax2+bx+c=0 , we can use the quadratic formula x=−b±b2−4ac2a

  x=−(−3)±(−3)2−4(1)(−1)2

Find the two values of x by simplifying further

    x=3+132x≈3.30277564x=3−132x≈−0.30277564

Check for extraneous solution by substituting the result in the original equation.

  x≈−0.30277564 makes the original equation FALSE.

  x≈3.30277564 makes the original equation TRUE.

Conclusion:

The solution to the equation ln(x+1)−ln(x−2)=lnx is        x≈3.303    _

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Chapter 3.4, Problem 59E

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
Ch. 3.1 - Prob. 11ECh. 3.1 - Prob. 12ECh. 3.1 - Prob. 13ECh. 3.1 - Prob. 14ECh. 3.1 - Prob. 15ECh. 3.1 - Prob. 16ECh. 3.1 - Prob. 17ECh. 3.1 - Prob. 18ECh. 3.1 - Prob. 19ECh. 3.1 - Prob. 20ECh. 3.1 - Prob. 21ECh. 3.1 - Prob. 22ECh. 3.1 - Prob. 23ECh. 3.1 - Prob. 24ECh. 3.1 - Prob. 25ECh. 3.1 - Prob. 26ECh. 3.1 - Prob. 27ECh. 3.1 - Prob. 28ECh. 3.1 - Prob. 29ECh. 3.1 - Prob. 30ECh. 3.1 - Prob. 31ECh. 3.1 - Prob. 32ECh. 3.1 - Prob. 33ECh. 3.1 - Prob. 34ECh. 3.1 - Prob. 35ECh. 3.1 - Prob. 36ECh. 3.1 - Prob. 37ECh. 3.1 - Prob. 38ECh. 3.1 - Prob. 39ECh. 3.1 - Prob. 40ECh. 3.1 - Prob. 41ECh. 3.1 - Prob. 42ECh. 3.1 - Prob. 43ECh. 3.1 - Prob. 44ECh. 3.1 - Prob. 45ECh. 3.1 - Prob. 46ECh. 3.1 - Prob. 47ECh. 3.1 - Prob. 48ECh. 3.1 - Prob. 49ECh. 3.1 - Prob. 50ECh. 3.1 - Prob. 51ECh. 3.1 - Prob. 52ECh. 3.1 - Prob. 53ECh. 3.1 - Prob. 54ECh. 3.1 - Prob. 55ECh. 3.1 - Prob. 56ECh. 3.1 - Prob. 57ECh. 3.1 - Prob. 58ECh. 3.1 - Prob. 59ECh. 3.1 - Prob. 60ECh. 3.1 - Prob. 61ECh. 3.1 - Prob. 62ECh. 3.1 - Prob. 63ECh. 3.1 - 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