College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.5 - Exponential and Logarithmic Equations - 4.5 Exercises - Page 447: 78


$x=\{ -2,2 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log_2 \sqrt{2x^2}=\dfrac{3}{2} ,$ convert to the equivalent exponential form. Then square both sides and use the properties of equality to isolate $x^2.$ Then take the square root of both sides. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \sqrt{2x^2}=2^{\frac{3}{2}} .\end{array} Squaring both sides results to \begin{array}{l}\require{cancel} (\sqrt{2x^2})^2=\left( 2^{\frac{3}{2}} \right)^2 \\\\ 2x^2=2^3 \\\\ 2x^2=8 \\\\ x^2=\dfrac{8}{2} \\\\ x^2=4 .\end{array} Taking the square root of both sides results to \begin{array}{l}\require{cancel} x=\pm\sqrt{4} \\\\ x=\pm2 .\end{array} Upon checking, both $ x=\{ -2,2 \} ,$ satisfy the original equation.
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