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: 13

Answer

$x\approx 3.240$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 6^{x+1}=4^{2x-1} ,$ take the logarithm of both sides. Then use the properties of logarithms and the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log6^{x+1}=\log4^{2x-1} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent to \begin{array}{l}\require{cancel} (x+1)\log6=(2x-1)\log4 .\end{array} Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(\log6)+1(\log6)=2x(\log4)-1(\log4) \\\\ x\log6+\log6=2x\log4-\log4 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} x\log6-2x\log4=-\log4-\log6 \\\\ x(\log6-2\log4)=-\log4-\log6 \\\\ x=\dfrac{-\log4-\log6}{\log6-2\log4} \\\\ x\approx 3.240 .\end{array}
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