College Algebra (11th Edition)

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

Chapter 4 - Section 4.3 - Logarithmic Functions - Summary Exercises on Inverse, Exponential, and Logarithmic Functions - Page 427: 41



Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2^x=\log_2 16 ,$ use the properties of logarithms to simplify the right side. Then express both sides in the same base and equate the exponents. $\bf{\text{Solution Details:}}$ Using exponents, the equation above is equivalent to \begin{array}{l}\require{cancel} 2^x=\log_2 2^4 .\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} 2^x=4\log_2 2 .\end{array} Since $\log_b b =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2^x=4(1) \\\\ 2^x=4 \\\\ 2^x=2^2 .\end{array} Since the bases are the same, then the exponents can be equated. That is, \begin{array}{l}\require{cancel} x=2 .\end{array}
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