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: 56

Answer

no solution

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \log(x^2+10x-39)-\log(x-3)=\log10 ,$ use the properties of logarithms to simplify the left-hand expression. Then drop the logarithm on both sides. Use properties of equality and concepts on solving quadratic equations. 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^2+10x-39}{x-3}=\log10 .\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^2+10x-39}{x-3}=10 .\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^2+10x-39}{x-3}=\dfrac{10}{1} \\\\ (x^2+10x-39)(1)=(x-3)(10) \\\\ x^2+10x-39=10x-30 .\end{array} Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} x^2=(10x-10x)+(-30+39) \\\\ x^2=9 .\end{array} Taking the square root of both sides (Square Root Principle), the solutions are \begin{array}{l}\require{cancel} x=\pm\sqrt{9} \\\\ x=\pm3 .\end{array} If $ x=3 ,$ the part of the given expression, $ \log(x-3) ,$ becomes $ \log (0) .$ This is not allowed since $ \log x$ is defined only for positive values of $x.$ If $ x=-3 ,$ the part of the given expression, $ \log(x-3) ,$ becomes $ \log (-6) .$ 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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