## Calculus (3rd Edition)

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