Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 60



Work Step by Step

$y=5^{x-2}$ $lny=ln5^{x-2}$ $=(x-2)ln5$ $lny=xln5-2ln5$ $\frac{dy}{y}=(ln5)dx$ $\frac{dy}{dx}=(ln5)y$ $=(ln5)e^{x-2}$ at (2,1), $\frac{dy}{dx}=(ln5)5^{2-2}$ $\frac{dy}{dx}=M=ln5$ Equation of Tangent: $(y-1)=ln5(x-2)$ $y-1=xln5-2ln5$ $y=(\ln5)x-2\ln5+1$
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