Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 389: 84

Answer

$y = \frac{a}{e^a - 1} (x - 1)$

Work Step by Step

The question asks for the equation of the line that passes through $(e^a, \ln (e^a)$ and the x-intercept of $\ln x$ Thus, at $x= e^a$, the y-coordinate would be equal to $\ln (e^a) = a$ The x-intercept of $\ln x$ is at (1, 0) So the slope would be $\frac {a - 0} {e^a - 1}$ = $\frac{a}{e^a - 1}$ Thus we have $y = \frac{a}{e^a - 1} x + b$ Substitute (1,0) into the equation $0 = \frac{a}{e^a - 1} (1) + b$ Thus $ b = -\frac{a}{e^a - 1}$ So $y = \frac{a}{e^a - 1} (x - 1)$
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