College Algebra (11th Edition)

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

Chapter 4 - Review Exercises: 40

Answer

$x\approx2.386$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ e^{x-1}=4 ,$ take the natural logarithm of both sides. Then use the laws of logarithms and the properties of equality to isolate the variable. Express the answer with $3$ decimal places. $\bf{\text{Solution Details:}}$ Taking the natural logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \ln e^{x-1}=\ln4 .\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} (x-1)\ln e=\ln4 .\end{array} Since $\ln e =1,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (x-1)(1)=\ln4 \\\\ x-1=\ln4 .\end{array} Using the properties of equality to isolate the variable results to \begin{array}{l}\require{cancel} x=1+\ln4 \\\\ x\approx2.386 .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.