Answer
$$y =-\frac{1}{8 \ln 2}(x-4)+1$$
Work Step by Step
Given $$y=\log _{2}\left(1+4 x^{-1}\right), \quad x=4$$
Rewrite $f(x) $ as
$$f(x)=\log _{2}\left(1+4 x^{-1}\right)=\frac{\ln \left(1+4 x^{-1}\right)}{\ln 2}$$
Since at $x=4$, $f(x) =1$ and
$$f'(x)= \frac{1}{\ln 2} \cdot \frac{1}{\left(1+4 x^{-1}\right)} \left(-4 x^{-2}\right)= -\frac{4}{x^{2}(\ln 2)\left(1+4 x^{-1}\right)} $$
Then $ m= f'(x)\bigg|_{x=4}=\dfrac{-1}{8 \ln 2} $
Hence, the tangent line is given by
\begin{align*}
\frac{y-y_1}{x- x_1}&=m\\
\frac{y-1 }{x- 4}&=\frac{-1}{8 \ln 2} \\
y&=-\frac{1}{8 \ln 2}(x-4)+1
\end{align*}
