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

Solve this equation




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







??

I think you mean:



In which case, multiply through by :



Which means that

Originally Posted by Fabio010

Solve this equation

LaTeX tip:
When an exponent has more that one character use {}.
[TEX]e^{3x}-2e^{-x} = 0[/TEX] gives

But when i write the function



it is different of the function




The solution is the same for y = 0

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

You started with

When I multiplied through by , it became:



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

Hum ok thanks

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

Thanks for the help.

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. is not the solution. From where I left off,

Taking natural logs of both sides:

and

Which makes the solution...

Originally Posted by Fabio010

Solve this equation




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

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

This does not follow at all!

??

Did you not consider checking that answer? If x= 0 then both and are
Do you beieve that 1- 2= 0?