1. ## solve for X

Solve for X: e^x(6-e^X)=9

2. Originally Posted by VNVeteran
Solve for X: e^x(6-e^X)=9
Distributing and rearranging, we get $e^{2x}-6e^x+9=0$.

3. From where Chris left off consider a substitution

$a = e^x$

giving

$a^{2}-6a+9=0$

Now you only need to solve a qudartic

4. Originally Posted by pickslides
From where Chris left off consider a substitution

$a = e^x$

giving

$a^{2}-6a+9=0$

Now you only need to solve a qudartic
(a-3) (a-3)
a=3 a=3
now what?

5. Originally Posted by VNVeteran
(a-3) (a-3)
a=3 a=3
now what?
$a=e^x$, so $e^x=3\implies x=\ldots$

6. it factors to
$
(e^x - 3)(e^x - 3)$

$e^x = 3$

take $ln$ to both sides

$lne^x = ln3$

$lne$ cancels out

so you have $x = ln3$