Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 8 - Section 8.4 - Multiplicative Inverses of Matrices and Matrix Equations - Exercise Set - Page 934: 94



Work Step by Step

Consider the given expression: $\log \left( x+4 \right)-\log \left( x-2 \right)=\log x$ Using the product identity, we get ${{\log }_{a}}A+{{\log }_{b}}B={{\log }_{ab}}(AB)$ So, $\begin{align} & \log \left( x+4 \right)-\log (x-2)=\log x \\ & \log \left( x+4 \right)=\log x+\log \left( x-2 \right) \\ & \log \left( x+4 \right)=\log \left( x\left( x-2 \right) \right) \end{align}$ Use anti log. Then, $\begin{align} & \left( x+4 \right)={{x}^{2}}-2x \\ & {{x}^{2}}-2x=x+4 \\ & {{x}^{2}}-2x-x-4=0 \\ & {{x}^{2}}-3x-4=0 \end{align}$ This equation can be solved by the quadratic equation of the form $a{{x}^{2}}+bx+c=0$ $\begin{align} & \frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & a=1,b=-3,c=-4 \\ & {{x}_{1,2}}=\frac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1} \\ & x=4,x=-1 \end{align}$ We eliminate the $-1$ solution because we can not take the log of a negative number.
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