Precalculus with Limits
Precalculus with Limits
3rd Edition
ISBN: 9781133947202
Author: Ron Larson
Publisher: Cengage Learning
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Chapter 3, Problem 101RE
To determine

Find the solution of the given equation using graphing utility and verify the result algebraically.

Expert Solution & Answer
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Answer to Problem 101RE

The value of the given equation is x≈1.482 .

Explanation of Solution

Given information:

The given equationis 2ln(x+3)−3=0 .

Calculation:

Use a graphing utility to graph the equation y=2ln(x+3)−3 and find the zeros.

The graph of the equation is given below.

  Precalculus with Limits, Chapter 3, Problem 101RE

From the graph, the value of the equation is x=1.482 .

Algebraic solution.

  2ln(x+3)−3=0⇒2ln(x+3)−3+3=0+3⇒2ln(x+3)=3⇒22ln(x+3)=32⇒22ln(x+3)=32⇒ln(x+3)=32⇒x+3=e32⇒x+3=4.481689⇒x+3−3=4.481689−3⇒x≈1.482

Hence thevalue of the given equationis x≈1.482 .

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Chapter 3, Problem 100RE
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Chapter 3, Problem 102RE

Chapter 3 Solutions

Precalculus with 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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