Exponential Functions

[Christina82]

I have done these before, however I have forgotten :roll:

I will attempt to type the problem, if someone knows how to do this, please let me know.

-2xe^-7x + x^2(e^-7x)=0

The answer to this is (0,-9) and I'm not sure how this is....

Thanks,
Christina

 

[Unco]

It isn't clear what you are trying to do with it nor what the answer refers to.

The equation \(\displaystyle \mbox{ -2xe^{-7x} + x^2e^{-7x} = 0}\) has solutions \(\displaystyle \mbox{x=0, 2}\) if that is the intention.

 

[galactus]

Is this your problem?:

\(\displaystyle {-2}xe^{-7x}+x^{2}e^{-7x}=0\)

\(\displaystyle e^{-7x}(x^{2}-2x)=0\)

It's easy to see that 0 and 2 are the answers.

Let's enter in -9 and see:

\(\displaystyle e^{-7(-9)}((-9)^{2}-2(-9))=99e^{63}\)

 

[Christina82]

Oh, I'm sorry. I wrote the wrong problem. However, I am unclear as to how you got that answer. I need to know the steps. I can use that problem, I just didn't have the answer to it...

Thanks for your help!

 

[galactus]

What makes \(\displaystyle x^{2}-2x=0\)?

0 and 2. Right?.

 

[Christina82]

Okay...

I understand that one. I think I'm confusing myself.

Here is the problem that I originally wanted to post w/ the answer.

I'm sorry about the typing on these, is there a math thing in here?

9xe^-7x + x^2(e^-7x)=0

This is the one that I do not understand how the answer is x=(0,-9) Maybe I am looking for something more complicated than it is....[/list]

 

[galactus]

This problem is essentially the same as the other.

Factor out \(\displaystyle e^{-7x}\)

\(\displaystyle e^{-7x}(x^{2}+9x)=0\)

\(\displaystyle x(x+9)=0\)

See how they get -9 and 0?.

Yes, there is a math thing in here. It's called LaTex. Click on quote in the upper right hand corner of the post to see how it was done. You can also learn about it by going to Forum Help at the top of the page.

 

[Christina82]

Thank you

I do now :O) Thanks! I can't believe I have forgotten a lot of this stuff! Thanks again!

Christina