Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.3 Logarithms and Their Derivatives - Exercises - Page 343: 61


$$R=\frac{12}{25 \ln 5}(z-3)+2$$

Work Step by Step

Given $$R(z)=\log _{5}\left(2 z^{2}+7\right), \quad z=3$$ Rewrite $R(z) $ as $$R(z)=\log _{5}\left(2 z^{2}+7\right)=\frac{\ln(2 z^{2}+7)}{\ln 5}$$ Since at $z=3$, $R(z) = \dfrac{\ln(25)}{\ln 5}=2$ and $$R'(z)= \frac{4z}{(2 z^{2}+7)\ln 5}$$ Then $ m= R'(z)\bigg|_{z=3}=\dfrac{4z}{(2 z^{2}+7)\ln 5}=\dfrac{12}{25 \ln 5}$ Hence, the tangent line is given by \begin{align*} \frac{R-R_1}{z- z_1}&=m\\ \frac{R- 2 }{z-3 }&=\frac{12}{25 \ln 5} \\ R &=\frac{12}{25 \ln 5}(z-3)+2 \end{align*}
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