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: 65


no solution

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log (9x+5)=3+\log (x+2) ,$ use the properties of logarithms to combine the logarithmic expressions. Then convert to exponential form. Use the properties of equality and concepts on solving quadratic equations. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \log (9x+5)-\log (x+2)=3 .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log \dfrac{9x+5}{x+2}=3 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} \log_{10} \dfrac{9x+5}{x+2}=3 \\\\ \dfrac{9x+5}{x+2}=10^3 \\\\ \dfrac{9x+5}{x+2}=1000 .\end{array} Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{9x+5}{x+2}=\dfrac{1000}{1} \\\\ (9x+5)(1)=(x+2)(1000) \\\\ 9x+5=1000x+2000 \\\\ 9x-1000x=2000-5 \\\\ -991x=1995 \\\\ x=\dfrac{1995}{-991} \\\\ x=-\dfrac{1995}{991} .\end{array} If $ x=-\dfrac{1995}{991} ,$ the part of the given expression, $ \log (x+2) ,$ becomes $ \log \left(-\dfrac{13}{991}\right) .$ This is not allowed since $ \log x$ is defined only for nonnegative values of $x.$ Hence, there is $\text{ no solution }.$
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