# Solving e^{3x}-2e^{-x} = 0.

• Nov 13th 2011, 05:28 AM
Fabio010
Solving e^{3x}-2e^{-x} = 0.
Solve this equation

$\displaystyle e^{3x}-2e^{-x} = 0$

I solved in this way, but i dont know if it is correct.

$\displaystyle e^{3x}-2e^{-x} = 0$

$\displaystyle e^{3x} = e^{-x} + e^{-x}$

$\displaystyle 3x + 2x = 0$

$\displaystyle x = 0$??
• Nov 13th 2011, 05:32 AM
Quacky
Re: Solving this equation ?
I think you mean:

$\displaystyle e^{3x}-2e^{-x}=0$

In which case, multiply through by $\displaystyle e^x$:

$\displaystyle e^{4x}-2=0$

Which means that $\displaystyle e^{4x}=2$
• Nov 13th 2011, 05:39 AM
Plato
Re: Solving this equation ?
Quote:

Originally Posted by Fabio010
Solve this equation
$\displaystyle e^(3x)-2e^(-x) = 0$

LaTeX tip:
When an exponent has more that one character use {}.
[TEX]e^{3x}-2e^{-x} = 0[/TEX] gives $\displaystyle e^{3x}-2e^{-x} = 0$
• Nov 13th 2011, 05:42 AM
Fabio010
Re: Solving this equation ?
But when i write the function

$\displaystyle e^{4x} = 2$

it is different of the function

$\displaystyle e^{3x} -2e^{-x} = 0$

The solution is the same for y = 0

but for other numbers like y = -20 it is different
• Nov 13th 2011, 05:58 AM
Quacky
Re: Solving this equation ?
You started with $\displaystyle e^{{\color{red}3}x}-2e^{-x}=0$

When I multiplied through by $\displaystyle e^x$, it became:

$\displaystyle e^{{\color{red}4}x}-2=0$

The functions are not the same, because I've multiplied through by $\displaystyle e^x$ which is variable. But when you multiply 0 by $\displaystyle e^x$, the value is still 0 - nothing changes, so the intercept doesn't change.
• Nov 13th 2011, 06:07 AM
Fabio010
Re: Solving this equation ?
Hum ok thanks

I understood now.. they are the same when x = 0 because e^0 = 1

Thanks for the help.
• Nov 13th 2011, 06:11 AM
Quacky
Re: Solving this equation ?
Quote:

Originally Posted by Fabio010
Hum ok thanks

I understood now.. they are the same when x = 0 because e^0 = 1

Thanks for the help.

I'm not sure that's the point I was making. $\displaystyle x=0$ is not the solution. From where I left off, $\displaystyle e^{4x}=2$

Taking natural logs of both sides:

$\displaystyle 4x\cdot{ln(e)}=ln(2)$ and $\displaystyle ln(e)=1$

Which makes the solution...
• Nov 13th 2011, 07:46 AM
HallsofIvy
Re: Solving this equation ?
Quote:

Originally Posted by Fabio010
Solve this equation

$\displaystyle e^{3x}-2e^{-x} = 0$

I solved in this way, but i dont know if it is correct.

$\displaystyle e^{3x}-2e^{-x} = 0$

$\displaystyle e^{3x} = e^{-x} + e^{-x}$

This is correct but it's not clear that this helps.

Quote:

$\displaystyle 3x + 2x = 0$
This does not follow at all!

Quote:

$\displaystyle x = 0$??
Did you not consider checking that answer? If x= 0 then both $\displaystyle e^{3x}$ and $\displaystyle e^x$ are $\displaystyle e^0= 1$
Do you beieve that 1- 2= 0?