Solve this equation

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

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

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

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

$\displaystyle 3x + 2x = 0$

$\displaystyle x = 0$??

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- Nov 13th 2011, 05:28 AMFabio010Solving 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 AMQuackyRe: 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 AMPlatoRe: Solving this equation ?
- Nov 13th 2011, 05:42 AMFabio010Re: 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 AMQuackyRe: 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 AMFabio010Re: 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 AMQuackyRe: Solving this equation ?
- Nov 13th 2011, 07:46 AMHallsofIvyRe: Solving this equation ?
This is correct but it's not clear that this helps.

Quote:

$\displaystyle 3x + 2x = 0$

Quote:

$\displaystyle x = 0$??

**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?