Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.6 Solving Exponential Equations and Logarithmic Equations - 12.6 Exercise Set: 23

Answer

$x\approx2.810$

Work Step by Step

Using the properties of equality, the given equation, $ 4.9^x-87=0 ,$ is equivalent to \begin{array}{l}\require{cancel} 4.9^x=87 .\end{array} Taking the natural logarithm of both sides and using the properties of logarithms, the value of the variable that satisfies the equation, $ 4.9^x=87 ,$ is \begin{array}{l}\require{cancel} \ln4.9^x=\ln87 \\\\ x\ln4.9=\ln87 \\\\ x=\dfrac{\ln87}{\ln4.9} \\\\ x\approx2.810 .\end{array}
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