Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 5 - 5.4 - Exponential and Logarithmic Equations - 5.4 Exercises - Page 396: 69

Answer

Graphically: $x=1.482$ Algebraically: $x=1.482$

Work Step by Step

Let $y_1=2\ln (x+3)$ and $y_2=x$. The graph using graphing utility is as shown. The intersection point is at $(1.482,3)$. Thus, the solution is $x=1.482$. Solving algebraically, divide both sides of the equation by $2$: $$2\ln (x+3)=3$$ $$\ln (x+3)=\frac{3}{2}$$ Rewriting the equation in exponential form where $\ln (x)=y$ is equivalent to $x=e^y$: $$x+3=e^{\frac{3}{2}}$$ $$x=e^{\frac{3}{2}}-3$$ $$x=1.482$$ Thus, the solution is verified algebraically.
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