Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 7 - Exponential and Logarithmic Functions - 7-6 Natural Logarithms - Practice and Problem-Solving Exercises - Page 481: 23


$x=\pm \sqrt{e^{4.9}} \approx \pm 11.588$

Work Step by Step

Subtract $1.1$ from each side to obtain: \begin{align*} 1.1+\ln{x^2}-1.1&=6-1.1\\ \ln{x^2}&=4.9 \end{align*} Recall: $$\ln{a} =y \longleftrightarrow e^y=a$$ Use the rule above to obtain: \begin{align*} \ln{x^2}&=4.9\\ e^{4.9}&=x^2 \end{align*} Take the square root of both sides: \begin{align*} \pm\sqrt{e^{4.9}}&=x\\ \end{align*} Use a calculator to obtain: $$x\approx \pm11.588$$
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