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

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

Solve this equation

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

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

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

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

$3x + 2x = 0$

$x = 0$??

2. ## Re: Solving this equation ?

I think you mean:

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

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

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

Which means that $e^{4x}=2$

3. ## Re: Solving this equation ?

Originally Posted by Fabio010
Solve this equation
$e^(3x)-2e^(-x) = 0$
LaTeX tip:
When an exponent has more that one character use {}.
[TEX]e^{3x}-2e^{-x} = 0[/TEX] gives $e^{3x}-2e^{-x} = 0$

4. ## Re: Solving this equation ?

But when i write the function

$e^{4x} = 2$

it is different of the function

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

The solution is the same for y = 0

but for other numbers like y = -20 it is different

5. ## Re: Solving this equation ?

You started with $e^{{\color{red}3}x}-2e^{-x}=0$

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

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

The functions are not the same, because I've multiplied through by $e^x$ which is variable. But when you multiply 0 by $e^x$, the value is still 0 - nothing changes, so the intercept doesn't change.

6. ## 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.

7. ## Re: Solving this equation ?

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. $x=0$ is not the solution. From where I left off, $e^{4x}=2$

Taking natural logs of both sides:

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

Which makes the solution...

8. ## Re: Solving this equation ?

Originally Posted by Fabio010
Solve this equation

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

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

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

$e^{3x} = e^{-x} + e^{-x}$
This is correct but it's not clear that this helps.

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

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