College Algebra (11th Edition)

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

Chapter 4 - Review Exercises - Page 469: 46

Answer

$x=2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ e^{8x}\cdot e^{2x}=e^{20} ,$ use the laws of exponents to simplify the left side. Then drop the bases on both sides and equate the exponents. Use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the equation above is equivalent to \begin{array}{l}\require{cancel} e^{8x+2x}=e^{20} \\\\ e^{10x}=e^{20} .\end{array} Since the bases are the same, drop the bases and equate the exponents. That is, \begin{array}{l}\require{cancel} 10x=20 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x=\dfrac{20}{10} \\\\ x=2 .\end{array}
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