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: 39

Answer

$x=2-10^{0.5}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log(2-x)=0.5 ,$ convert to exponential form. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $\log x=\log_{10} x,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \log_{10}(2-x)=0.5 .\end{array} Since $y=b^x$ is equivalent to $\log_b y=x,$ the exponential form of the equation above is \begin{array}{l}\require{cancel} 10^{0.5}=2-x .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} x=2-10^{0.5} .\end{array}
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