# Solve for x

• February 26th 2006, 12:32 AM
totalnewbie
Solve for x
How to solve for x ?
• February 26th 2006, 04:06 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
How to solve for x ?

You should check the question, as set there is no solution for
real $x$ (sketch the curve for $x>0$ to see why).

Problem was: Solve

$
\ln(x)=\frac{3x^2}{2 \mathrm{e}}
$

RonL
• February 26th 2006, 04:57 AM
totalnewbie
It seems to me that I have lost. My goal was to find common tangent for two functions: f(x)=x^2/2e and g(x)=ln(x)
It was also said that x-coordinate is sqrt(e^1) for one tangent point.
• February 26th 2006, 05:24 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
It seems to me that I have lost. My goal was to find common tangent for two functions: f(x)=x^2/2e and g(x)=ln(x)
It was also said that x-coordinate is sqrt(e^1) for one tangent point.

Some trial and error might help here.

Trial 1
Assume the $x=\sqrt{e}$ point is the tangency point on $f(x)$
Then we can find the slope of the tangent at $x=\sqrt{e}$, and
the corresponding value of $f(x)$, which will allow us to find
the equation of the tangent. Now check that this line is a tangent
to $g(x)$, if it is our job is done, if not proceed to Trial 2

Trial 2
Assume the $x=\sqrt{e}$ point is the tangency point on $g(x)$,
then proceed as in Trial 1, but with the roles of $f$ and $g$
interchanged.

RonL
• February 26th 2006, 07:06 AM
totalnewbie
I have tried it.
The green line doesn't touch the red parabola. But it should touch.
• February 26th 2006, 07:12 AM
earboth
Quote:

Originally Posted by totalnewbie
It seems to me that I have lost. My goal was to find common tangent for two functions: f(x)=x^2/2e and g(x)=ln(x)
It was also said that x-coordinate is sqrt(e^1) for one tangent point.

Hello,

if there exists a common tangent then the gradient of both functions must be equal. So first calculate the drivative of both functions:
$\frac{df}{dx}=\frac{1}{e} \cdot x$
$\frac{dg}{dx}=\frac{1}{x}$

Both are equal: $\frac{1}{e} \cdot x=\frac{1}{x}$

Solve for x and you'll get: $x=\sqrt{e}$

Complete the coordinates:
$f(\sqrt{e})=\frac{1}{2}$
$g(\sqrt{e})=\frac{1}{2}$

That means that the graphs of both functions have one common point: the tangent point. There exists one tangent (that's a special case, normally there must be two tangents!)

Use the point-slope-formula: $t: y=\frac{1}{\sqrt{e}} \cdot (x-\sqrt{e})+\frac{1}{2}$

I've attached a drawing to demonstrate my results.

Greetings

EB
• February 26th 2006, 07:20 AM
earboth
Quote:

Originally Posted by totalnewbie
I have tried it.
The green line doesn't touch the red parabola. But it should touch.

Hello,

you have typed the function f like this: x^2/2e. Then your computer will calculate $\frac{x^2}{2} \cdot e$.

You have to use parantheses: $\frac{x^2}{(2 \cdot e)}$

Good luck.

Greetings

EB
• February 26th 2006, 08:41 AM
CaptainBlack
Quote:

Originally Posted by totalnewbie
I have tried it.
The green line doesn't touch the red parabola. But it should touch.

So now switch to the other case (where the tangent at $x=\sqrt{e}$ is to the other curve).

RonL