# A new puzzle?

• October 18th 2010, 10:36 AM
wonderboy1953
A new puzzle?
Some days ago, I did a puzzle using exp(x) and ln(x) and it occurred to me that these functions can be used to start off a different type of puzzle.

Let x = 1. Then you have the points (1,e) and (1,0). Now draw tangents to the curves at those points. The tangent lines (most likely - I haven't formally figured it out this far) would intersect at a point. Figure out what this point is. Now let x = 2 to get the points (2, exp(2)) and (2, ln(2)). Now draw tangents to those two points and figure out what the point of intersection is for the next tangent lines. Let x take on various different values and derive more points of intersection.

Can you produce a graph showing the locus of intersection points? As an extra bonus question, can you derive an equation for these locus points? (to take it a step further, you may want to explore what happens letting y = x +1 or using parabolas for variety sakes).
• October 23rd 2010, 12:29 PM
Ackbeet
This was a fun problem. Here's my answer. For the function $e^{x},$ we write the equation of the tangent line to the point $(x,e^{x})$ as $y_{1}(x_{1}).$ Similarly, for the function $\ln(x),$ we write the equation of the tangent line as $y_{2}(x_{2}).$ Now, given that $(e^{x})'=e^{x}$ and $(\ln(x))'=1/x,$ we can immediately write down the equations of the tangent lines:

$y_{1}-e^{x}=e^{x}(x_{1}-x),$ and
$y_{2}-\ln(x)=\dfrac{1}{x}(x_{1}-x).$

We want to set these equal. Obviously, this will have to happen when $x_{1}=x_{2},$ so re-define $\xi=x_{1}=x_{2},$ and set

$e^{x}+e^{x}(\xi-x)=\ln(x)+\dfrac{1}{x}(\xi-x).$

Solving for $\xi-x$ yields

$\xi-x=\dfrac{x(e^{x}-\ln(x))}{1-xe^{x}}.$

Plugging this back into either $y_{1}$ or $y_{2}$ (it shouldn't matter, although I did it in both as a check, and they both come out the same), you get

$y=\dfrac{e^{x}(1-x\ln(x))}{1-xe^{x}}.$

This is valid when $xe^{x}\not=1.$