Calculus (3rd Edition)

$$y =-\frac{1}{8 \ln 2}(x-4)+1$$
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*}