# How to solve this simple equation involving an exponential function?

• May 21st 2011, 05:49 AM
Yehia
How to solve this simple equation involving an exponential function?
So given these two functions:

f(x) = e^(-2x + 3)
g(x) = √(ax)

how can you find the value of a, given that the two curves intersect perpendicularly??
the way i see it is that you have to many unknowns (i equated the derivatives).

Help much appreciated!!! thanks!!!
• May 21st 2011, 06:37 AM
topsquark
Quote:

Originally Posted by Yehia
So given these two functions:

f(x) = e^(-2x + 3)
g(x) = √(ax)

how can you find the value of a, given that the two curves intersect perpendicularly??
the way i see it is that you have to many unknowns (i equated the derivatives).

Help much appreciated!!! thanks!!!

The solution to the intersection point can only be found numerically so it is next to impossible to find an expression for a that way. However we can be tricky.

First we are looking for a point x such that f(x) = g(x). That is we have a point x = b on the functions f(x) and g(x) such that
$e^{-2b + 3} = \sqrt{ab}$

Now, perpendicular slopes follow the rule that m' = -1/m. So we are looking for a at the intersection point such that
$g'(x) = -1/f'(x)$

Evaluate this at x = b and see if you can use the intersection condition to write this equation as a function of a only.

-Dan
• May 21st 2011, 06:47 AM
Yehia
Quote:

Originally Posted by topsquark
The solution to the intersection point can only be found numerically so it is next to impossible to find an expression for a that way. However we can be tricky.

First we are looking for a point x such that f(x) = g(x). That is we have a point x = b on the functions f(x) and g(x) such that
$e^{-2b + 3} = \sqrt{ab}$

Now, perpendicular slopes follow the rule that m' = -1/m. So we are looking for a at the intersection point such that
$g'(x) = -1/f'(x)$

Evaluate this at x = b and see if you can use the intersection condition to write this equation as a function of a only.

-Dan

Dan, I did EXACTLY that. i ended up with BOTH those equations, BUT both happened to be the same. can you please show me the way forward or at least do it yourself and see if you get an answer?
• May 21st 2011, 07:00 AM
topsquark
Quote:

Originally Posted by Yehia
Dan, I did EXACTLY that. i ended up with BOTH those equations, BUT both happened to be the same. can you please show me the way forward or at least do it yourself and see if you get an answer?

$f(x) = e^{-2x + 3} \implies f'(x) = -2e^{-2x + 3}$

$g(x) = \sqrt{ax} \implies g'(x) = \frac{\sqrt{a}}{2\sqrt{x}}$

At the intersection point we have f(b) = g(b), or
$e^{-2b + 3} = \sqrt{ab}$

Also we know that g'(b) = -1/f'(b), so
$\frac{\sqrt{a}}{2\sqrt{b}} = \frac{1}{2 e^{-2b + 3}}$

But what does the intersection point equation say that $e^{-2b + 3}$ is equal to?

-Dan
• May 21st 2011, 07:14 AM
Yehia
Quote:

Originally Posted by topsquark
$f(x) = e^{-2x + 3} \implies f'(x) = -2e^{-2x + 3}$

$g(x) = \sqrt{ax} \implies g'(x) = \frac{\sqrt{a}}{2\sqrt{x}}$

At the intersection point we have f(b) = g(b), or
$e^{-2b + 3} = \sqrt{ab}$

Also we know that g'(b) = -1/f'(b), so
$\frac{\sqrt{a}}{2\sqrt{b}} = \frac{1}{2 e^{-2b + 3}}$

But what does the intersection point equation say that $e^{-2b + 3}$ is equal to?

-Dan

Ok, i went on and now i end up with:

b^5 = e^(12-8b)
• May 21st 2011, 07:55 AM
topsquark
Quote:

Originally Posted by Yehia
Ok, i went on and now i end up with:

b^5 = e^(12-8b)

Please show us how you got that equation. That will help us pinpoint the error.

-Dan
• May 21st 2011, 08:24 AM
Yehia
Quote:

Originally Posted by topsquark
Please show us how you got that equation. That will help us pinpoint the error.

-Dan

no
• May 21st 2011, 09:04 AM
topsquark
Quote:

Originally Posted by Yehia
no

Then you no longer require help. Thread closed.

-Dan