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