# Simplifying E^x equation

• Mar 25th 2010, 09:33 AM
atljogger
Simplifying E^x equation
Can someone show me how to simplify an exponential function E^x when there is a variable in the exponent? I would like to get X out of the exponent and solve for it.

$y =(4X+3)\frac{1}{1+e^{3X-2}}
$

or more complicated
$y =(4X+3)\frac{1}{1+e^{3X^3+2x^2+4X-2}}
$

I was able to get some help on finding the derivative in the Calculus section but have been unable to figure out how to simplify this equation.

Also, how do I calculate the square of an e^x equation:
$y = {(1+e^{3X-2})}^2
$
• Mar 25th 2010, 09:57 AM
Hello atljogger
Quote:

Originally Posted by atljogger
Can someone show me how to simplify an exponential function E^x when there is a variable in the exponent? I would like to get X out of the exponent and solve for it.

$y =(4X+3)\frac{1}{1+e^{3X-2}}
$

or more complicated
$y =(4X+3)\frac{1}{1+e^{3X^3+2x^2+4X-2}}
$

I was able to get some help on finding the derivative in the Calculus section but have been unable to figure out how to simplify this equation.

Also, how do I calculate the square of an e^x equation:
$y = {(1+e^{3X-2})}^2
$

The bad news is that you'll only be able to solve equations like these by numerical methods - there won't be an analytical solution. So you'll have to settle for approximate answers.

The good news is that squaring the final expression you've written down is pretty easy. Just expand in the usual way, using $(a+b)^2 = a^2+2ab+b^2$. So:
${(1+e^{3X-2})}^2 = 1^2 + 2.1.e^{3X-2}+(e^{3X-2})^2$
$= 1 + 2e^{3X-2}+e^{6X-4}$
• Mar 25th 2010, 11:55 AM
atljogger
Thanks Grandad. Is there a way to solve for X since its in both the formula and the exponent? For example, I took a derivatve and want to solve for X by setting these 2 equations equal:

$
1+e^{3X-2}=(4X+3)*({3e^{3X-2})}$
• Mar 26th 2010, 05:19 AM
HallsofIvy
You would have to use the "Lambert W function" which is defined as the inverse function to $f(x)= xe^x$.
• Mar 26th 2010, 11:57 AM
atljogger
Thanks. I reduced a variation of this equation to this:

$ln(x) + .75x = 9.5
$

Can the Lambert Function be used to solve for X - the two terms with X are being added, not multiplied?
• Mar 26th 2010, 01:59 PM
HallsofIvy
Quote:

Originally Posted by atljogger
Thanks. I reduced a variation of this equation to this:

$ln(x) + .75x = 9.5
$

Can the Lambert Function be used to solve for X - the two terms with X are being added, not multiplied?

Sure. Rewrite that as ln(x)= 9.5- .75x and take the exponential of both sides: $x= e^{9.5- .75x}= e^{9.5}e^{-.75x}$.

Multiply both sides by $e^{.75x}$ to get $xe^{.75x}= e^{9.5}$ which is the same as $.75xe^{.75x}= .75e^{9.5}$[/tex]. Let y= .75x and your equation becomes $ye^y= .75e^{9.5}$. Now you can use the Lambert W function- $y= W(.75e^{9.5})$ Then, of course, $x= \frac{y}{.75}= \frac{W(.75e^{9.5})}{.75}$.