Exponential Problem: Solving (b-a-2)e^b + (b-a)e^a with Constant a and b

[zeithief]

i have an exponential problem that i could not solve.
$$(b-a-2)e^b + (b-a)e^a$$ where a and b are constant. Is it the same as $$b^2 -2e^b + 6b - a^2 + 2e^a - 6a$$ ?
Hope that someone kind could spend some time helping me and list down the steps for me. If they are not equal can you pls tell me too?

Thank you!

 

[whozum]

They aren't equal. Not sure what you are doing,
$$(b-a-2)e^b + (b-a)e^a$$

Just use regular old distribution,

$$(a+b+c+d)x = ax + bx + cx + dx$$

You shouldn't end up with anything squared, and everything should have an exponential attached to it.

 

[shyboy]

i have an exponential problem that i could not solve.
$$(b-a-2)e^b + (b-a)e^a$$ where a and b are constant. Is it the same as $$b^2 -2e^b + 6b - a^2 + 2e^a - 6a$$ ?
Hope that someone kind could spend some time helping me and list down the steps for me. If they are not equal can you pls tell me too?

Thank you!

sure, easy. if b=a=0 then the left side left gives you -2 and the right side gives you 0.

 

[Jameson]

This cannot be solved. There is no equation. That expression needs to equal something otherwise the only thing you can do is simplify it, but that seems to be done already.

Jameson

 

[Zurtex]

i have an exponential problem that i could not solve.
$$(b-a-2)e^b + (b-a)e^a$$ where a and b are constant. Is it the same as $$b^2 -2e^b + 6b - a^2 + 2e^a - 6a$$ ?
Hope that someone kind could spend some time helping me and list down the steps for me. If they are not equal can you pls tell me too?

Thank you!

If you mean you want to find an instance where:

$$(b-a-2)e^b + (b-a)e^a = b^2 -2e^b + 6b - a^2 + 2e^a - 6a$$

I don't think there is an algebraic way of doing it. I would like to note also that a = 0 and b = 0 is not a solution. Various mathematical programs will easily be able to approximate answers.