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: 55


no solution

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log(x-10)-\log(x-6)=\log2 ,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. Use the properties of equality to isolate the variable. Finally, do checking of the solution with the original equation. $\bf{\text{Solution Details:}}$ Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log\dfrac{x-10}{x-6}=\log2 .\end{array} Since the logarithm on both sides have the same base, then the logarithm can be dropped. Hence, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{x-10}{x-6}=2 .\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{x-10}{x-6}=\dfrac{2}{1} \\\\ (x-10)(1)=(x-6)(2) \\\\ x-10=2x-12 .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} x-10=2x-12 \\\\ x-2x=-12+10 \\\\ -x=-2 \\\\ x=2 .\end{array} If $ x=2 ,$ the part of the given expression, $ \log(x-10) ,$ becomes $ \log(-8) .$ This is not allowed since $ \log x$ is defined only for positive values of $x.$ Hence, there is $\text{ no solution .}$
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