Math Help - Solving Exponential Equations

1. Solving Exponential Equations

Hello,

I'm a new user and I was hoping someone can help me with this problem?

Edit: sorry I attached the wrong problem, it's actually:

x^2e^x+xe^2-e^x=0

hope this is clear

Thanks!

2. Hello l flipboi l

Welcome to Math Help Forum!
Originally Posted by l flipboi l
Hello,

I'm a new user and I was hoping someone can help me with this problem?

Sorry for the "crappy" illustration

[IMG]file:///Users/Mark/Library/Caches/TemporaryItems/moz-screenshot.png[/IMG][IMG]file:///Users/Mark/Library/Caches/TemporaryItems/moz-screenshot-1.png[/IMG]
Divide through by a factor $e^2$. Then use the quadratic formula on the resulting equation: $x^2+x-1=0$.

3. Originally Posted by l flipboi l
Hello,

I'm a new user and I was hoping someone can help me with this problem?

Edit: sorry I attached the wrong problem, it's actually:

x^2e^x+xe^2-e^x=0

hope this is clear

Thanks!
I doubt an algebraic solution can be found using elementary functions. Please state the entire question from which the equation has come from.

4. Sorry I keep messing up, the equation is:

x^2e^x+xe^x-e^x=0

What's confusing me is the e^x, I can't seem to get anywhere near a solution

Thanks,
-Mark

5. Originally Posted by l flipboi l
Sorry I keep messing up, the equation is:

x^2e^x+xe^x-e^x=0

What's confusing me is the e^x, I can't seem to get anywhere near a solution

Thanks,
-Mark
Originally Posted by Grandad

Divide through by a factor $e^2$. Then use the quadratic formula on the resulting equation: $x^2+x-1=0$.

To elaborate you can factor out $e^x$ as it's a common factor:

$x^2e^x+xe^x-e^x= e^x(x^2+x-1) = 0$

Dividing through by $e^x$ (this is allowed because for real x: $e^x > 0$) gives $x^2+x-1 = 0$

Use the quadratic formula to solve for x

6. Originally Posted by e^(i*pi)
To elaborate you can factor out $e^x$ as it's a common factor:
$x^2e^x+xe^x-e^x= e^x(x^2+x-1) = 0$
Dividing through by $e^x$ (this is allowed because for real x: $e^x > 0$) gives $x^2+x-1 = 0$