Find the tangent to the graph of the function $f (x) = \ln (x-1) ^ 2$ which is perpendicular to the line $p$: $y = \frac{2}{5}x +2$
I decided $kp = 2/5$ and planted in $kn = -1/kp = -5/2 \to kt = -5/2$. Then I derived the function $f'(x) = 2/(x-1)$.
Substituted into the equation $kt = -5/2 = 2/(x-1) \to x=1/5$. And put in the original function. And determined $y=\ln(16/25)$
Is this procedure correct and how do I express the tangent equation?
So we want to find a tangent to $f(x) = \ln ((x-1)^2)$ that has the slope of $\frac{-1}{2/5} = -\frac{5}{2}$. Indeed the derivative you got was correct, so we just need to find the point $(x, f(x))$ for which we have a tangent with the slope $-\frac{5}{2}$. So
$$ f'(x) = -\frac{5}{2} = \dfrac{2}{x-1} \implies x=\frac{1}{5} $$
So the point $\left(\frac{1}{5}, \ln \left(\frac{16}{25}\right)\right)$ has a tangent to $f(x)$ with the slope $-\frac{5}{2}$, So
$$ y - \ln \left(\frac{16}{25}\right) = -\frac{5}{2}\left(x-\frac{1}{5}\right) $$