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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!

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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 $\ln(xy)=2.5x$.

Mathematica graphics

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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:

$\quad \ln(xy) = 2.5x$

enter image description here

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Is this okay, now user59714? If you've found the answer helpful, you may accept it. To accept an answer, you need only click on the $\checkmark$ to the left of the answer. Plus, you get $2$ reputation points for every accepted answer! – amWhy Mar 6 '13 at 22:17

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