Where is the tangent line to this curve horizontal? Find the $x$-co-ordinate of the point where the tangent line to the curve $\ln(xy)=2.5x$ is horizontal.
So, using implicit differentiation, I got $\frac{dy}{dx}$ to be $y(2.5 - \frac{1}{x})$. What do I do next? Thanks!
 A: If you want a horizontal tangent, then you need to look for where ${dy\over dx}=0$.
Edit: As indicated, solve ${dy\over dx}=0\implies y(2.5-1/x)=0\implies y=0\text{ or }x={2\over 5}$. But $y=0$ is not in the domain of the original relation, $\ln(xy)=2.5x$, so the horizontal tangent occurs only when $x=2/5\implies y=5e/2$.
This is depicted graphically below: the dashed line is the horizontal tangent passing thru $(2/5,5e/2)$ on the graph of $\ln(xy)=2.5x$.

A: A tangent line (any line) is horizontal (parallel to the x-axis) when its slope equals $0$.
So you need to set $\;dy/dx = 0$:
When is $\;\dfrac{dy}{dx}= y\,(2.5 - \dfrac{1}{x}) = 0\;\;$? 
Clearly, $\dfrac{dy}{dx} = 0\;$ when $\;y = 0.\;$ 
And for what value of $\,x^*\,$ is $\;\dfrac{dy}{dx}= y\,\left(2.5 - \dfrac{1}{x^*}\right) = 0\;\;?\tag{$x^*$}$
Now, use the original equation $\;\;\ln(xy)=2.5x\;\;$ to solve for $\,x\,$ at $y = 0$: your curve is undefined there at $y = 0.\;\;$ As it is not in the domain of the function. What might be happening there? 
So next, use the value $\;x = (x^*)\,$ to solve for $\,y^*\,$, using the original equation. That point $\,(x^*, y^*)\,$ will be the one point at which the line tangent to your curve is horizontal. $$\;\;\ln(xy)=2.5x\;\;\iff \exp{(\ln(xy))} = \exp(2.5x)$$
$$\iff xy = \exp(\frac52 x) \iff y = \frac{\exp(\frac 52 x)}{x}$$ $$x=2/5\implies y=5e/2$$
Sometimes graphing helps to shed insight with respect to the behavior of a curve:
$\quad \ln(xy) = 2.5x$

