How to solve for x ?

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- Feb 26th 2006, 12:32 AMtotalnewbieSolve for x
How to solve for x ?

- Feb 26th 2006, 04:06 AMCaptainBlackQuote:

Originally Posted by**totalnewbie**

real $\displaystyle x$ (sketch the curve for $\displaystyle x>0$ to see why).

Problem was: Solve

$\displaystyle

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

$

RonL - Feb 26th 2006, 04:57 AMtotalnewbie
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. - Feb 26th 2006, 05:24 AMCaptainBlackQuote:

Originally Posted by**totalnewbie**

Trial 1

Assume the $\displaystyle x=\sqrt{e}$ point is the tangency point on $\displaystyle f(x)$

Then we can find the slope of the tangent at $\displaystyle x=\sqrt{e}$, and

the corresponding value of $\displaystyle f(x)$, which will allow us to find

the equation of the tangent. Now check that this line is a tangent

to $\displaystyle g(x)$, if it is our job is done, if not proceed to Trial 2

Trial 2

Assume the $\displaystyle x=\sqrt{e}$ point is the tangency point on $\displaystyle g(x)$,

then proceed as in Trial 1, but with the roles of $\displaystyle f$ and $\displaystyle g$

interchanged.

RonL - Feb 26th 2006, 07:06 AMtotalnewbie
I have tried it.

The green line doesn't touch the red parabola. But it should touch. - Feb 26th 2006, 07:12 AMearbothQuote:

Originally Posted by**totalnewbie**

if there exists a common tangent then the gradient of both functions must be equal. So first calculate the drivative of both functions:

$\displaystyle \frac{df}{dx}=\frac{1}{e} \cdot x$

$\displaystyle \frac{dg}{dx}=\frac{1}{x}$

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

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

Complete the coordinates:

$\displaystyle f(\sqrt{e})=\frac{1}{2}$

$\displaystyle g(\sqrt{e})=\frac{1}{2}$

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: $\displaystyle t: y=\frac{1}{\sqrt{e}} \cdot (x-\sqrt{e})+\frac{1}{2}$

I've attached a drawing to demonstrate my results.

Greetings

EB - Feb 26th 2006, 07:20 AMearbothQuote:

Originally Posted by**totalnewbie**

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

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

Good luck.

Greetings

EB - Feb 26th 2006, 08:41 AMCaptainBlackQuote:

Originally Posted by**totalnewbie**

RonL