College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Review Exercises: 45

Answer

$x\approx-2.487$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 6^{x-3}=3^{4x+1} ,$ take the logarithm of both sides. Use the laws of logarithms and the properties of equality to isolate the variable. Finally, express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log6^{x-3}=\log3^{4x+1} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} (x-3)\log6=(4x+1)\log3 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x(\log6)-3(\log6)=4x(\log3)+1(\log3) \\\\ x\log6-3\log6=4x\log3+\log3 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} -\log3-3\log6=4x\log3-x\log6 \\\\ -\log3-3\log6=x(4\log3-\log6) \\\\ \dfrac{-\log3-3\log6}{4\log3-\log6}=x \\\\ x=\dfrac{-\log3-3\log6}{4\log3-\log6} \\\\ x\approx-2.487 .\end{array}
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