College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 37

Answer

$x\approx1.09$

Work Step by Step

Find the natural log of both sides and then pull out the exponent to the left side. Then solve for $x$. $7^{x+2}=410$ $\ln(7^{x+2})=\ln(410)$ $(x+2)\ln(7)=\ln(410)$ $(x+2)=\frac{\ln(410)}{\ln(7)}$ $x=\frac{\ln(410)}{\ln(7)}-2$ $x\approx1.09$
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