Answer
$$f =\frac{8}{\ln 2}\left(w-\frac{1}{8}\right)-3$$
Work Step by Step
Given $$f(w)=\log _{2} w, \quad w=\frac{1}{8}$$
Rewrite $f(w) $ as
$$f(w)=\log _{2} w=\frac{\ln w}{\ln 2}$$
Since at $w=1/8$, $f(w) = \dfrac{\ln(1/8)}{\ln 2}= -3$ and
$$f'(w)= \frac{1}{w\ln 2}$$
Then $ m= f'(w)\bigg|_{w=1/8}= \dfrac{1}{w\ln 2}= \dfrac{8}{\ln 2} $
Hence, the tangent line is given by
\begin{align*}
\frac{f-f_1}{w- w_1}&=m\\
\frac{f+3 }{w-1/8 }&=\dfrac{8}{\ln 2} \\
f &=\frac{8}{\ln 2}\left(w-\frac{1}{8}\right)-3
\end{align*}
