In other words: you're asked to find a general equation of a line that is normal (perpendicular) to any given point of $$\ln x+C$$. The normal line is perpendicular to the tangent line, which means the slope of the normal line is the negative reciprocal of the slope of the tangent. The slope of the tangent is given by the derivative. $y=\ln x+C~~\Rightarrow~~\frac{dy}{dx}=\frac{1}{x}~~\Rightarrow~~\text{slope}_{\text{normal}}=-x$ So for some point $$(x_0,y_0)$$ on $$y=\ln x+C$$, that is, $$(x_0,\ln x_0+C)$$, the normal line has slope $$-x_0$$. Also, since it passes through $$x_0,\ln x_0+C$$, it will have the equation $y-\ln x_0-C=-x_0(x-x_0)~~\iff~~y=\ln x_0-x_0(x-x_0)+C$ I'm not sure where the $$k$$ comes in; I don't really have any context for the problem. This is just how I would approach it.