Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 6 - Logarithmic Functions - 6.6 Solving Logarithmic Equations - 6.6 Exercises - Page 535: 48

Answer

The correct answer is $x = \sqrt \frac{100}{3}$. The student added $x$ to $3x$, when instead, she should be multiplying $x$ by $3x.$ Instead of obtaining $4x$, she should be obtaining $3x^{2}$.

Work Step by Step

$\log x + \log 3x = 2$ $\log 3x^{2} = 2$ $\log_{10} 3x^{2} = 2$ $10^{2} = 3x^{2}$ $100 = 3x^{2}$ $x^{2} = \frac{100}{3}$ $x = ±\sqrt \frac{100}{3}$ $x = \sqrt \frac{100}{3}$ The student added $x$ to $3x$, when instead, she should be multiplying $x$ by $3x.$ Instead of obtaining $4x$, she should be obtaining $3x^{2}$.
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