Precalculus (6th Edition) Blitzer

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

Chapter 3 - Review Exercises - Page 513: 78


The required value is $ x=2$

Work Step by Step

Consider the given equation, $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x $ Apply the quotient rule of logarithms $\begin{align} & \ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x \\ & \ln \frac{\left( x+4 \right)}{\left( x+1 \right)}=\ln x \end{align}$ Comparing both sides, we get, $\begin{align} & \frac{\left( x+4 \right)}{\left( x+1 \right)}=x \\ & \left( x+4 \right)=x\left( x+1 \right) \\ & \left( x+4 \right)={{x}^{2}}+x \\ \end{align}$ Add to both sides $-x $ $\begin{align} & \left( x+4-x \right)={{x}^{2}}+x-x \\ & {{x}^{2}}=4 \end{align}$ Taking the square root on both sides, we get: $\begin{align} & \sqrt{{{x}^{2}}}=\sqrt{4} \\ & x=2,-2 \end{align}$ Here $ x=-2$ is not the solution of the given expression because $\begin{align} & \ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x \\ & \ln \left( -2+4 \right)-\ln \left( -2+1 \right)=\ln \left( -2 \right) \\ & \ln \left( 2 \right)-\ln \left( -1 \right)=\ln \left( -2 \right) \end{align}$ So, by definition of ${{\log }_{a}}b $, $ b>0$ but $-1$ and $-2$ is not greater than zero Thus, $ x=-2$ does not satisfy the equation $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x $ Hence, $ x=2$ is the only solution of the given expression. Therefore, the value of $\ln \left( x+4 \right)-\ln \left( x+1 \right)=\ln x $ is $ x=2$.
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