Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.5 Logarithmic Functions - 2.5 Exercises - Page 98: 59

Answer

$\ln6 +1$

Work Step by Step

$e^{k-1}=6$ $\ln e^{k-1}=\ln 6$ $(k-1) \ln e= \ln 6$ $k=\frac{\ln 6}{\ln e}+1 =\ln6 +1$
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