Finding the roots of this equation

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April 11th 2011, 12:46 PM
IanCarney

Finding the roots of this equation

$\displaystyle y=e^{-3x} - e^{2x}$

I factored it to: $\displaystyle y=(e^{-3x})(e^{2x}-1)$

$\displaystyle e^{-3x}$ Never equals zero (right?)

How would I solve for $\displaystyle e^{2x}-1=0$ ?

Edit: I know the answer is $x=0$ by looking at it, but how would I solve a similar equation?
April 11th 2011, 12:51 PM
Plato

Quote:

Originally Posted by IanCarney

$\displaystyle y=e^{-3x} - e^{2x}$
I factored it to: $\displaystyle y=(e^{-3x})(e^{2x}-1)$

The correct factorization is $e^{-3x}\left(1-e^{5x}\right)$ .
April 11th 2011, 12:54 PM
IanCarney

Quote:

Originally Posted by Plato

The correct factorization is $e^{-3x}\left(1-e^{5x}\right)$ .

Ah, yes. I wrote down the factored form of its derivative by accident >.<

Here's what I'm trying to solve:

$e^{2x}=3$

Would I take the log of both sides? Because I get a different answer.
April 11th 2011, 12:54 PM
TKHunny

Factor in the usual way. $\left(e^{x}+1\right)\cdot\left(e^{x}-1\right)=0$

You'll have trouble with the fist piece, for reasons similar to that 3x you already threw out.

You'll need a logarithm or an emperical rule for the last piece. Very informally, $Number^{What} = 1$ ?
April 11th 2011, 01:25 PM
e^(i*pi)

Quote:

Originally Posted by IanCarney

$e^{2x}=3$

Would I take the log of both sides? Because I get a different answer.

Natural Log: $2x = \ln(3)$
April 11th 2011, 08:10 PM
johnny

Quote:

Originally Posted by IanCarney

$\displaystyle y=e^{-3x} - e^{2x}$

I factored it to: $\displaystyle y=(e^{-3x})(e^{2x}-1)$

$\displaystyle e^{-3x}$ Never equals zero (right?)

How would I solve for $\displaystyle e^{2x}-1=0$ ?

Edit: I know the answer is $x=0$ by looking at it, but how would I solve a similar equation?

y = e^(-3x) - e^(2x) = e^(-3x) (1 - e^(5x)) = 0
1 - e^(5x) = 0
1 = e^(5x)
ln 1 = ln e^(5x) = 5x
ln 1 = 0
∴ x = 0/5 = 0

Moderator edit: Quote of OP has been added. This post answers the original question posted. However, it should be noted that in post #3 the original poster gave a new question.