Problem solving equation with negative exponent

[CentreShifter]

I'm trying to calculate maximum power. Once I computed power as a function of time I took its derivative and set it equal to zero. Now it need to solve for time and I can't seem to get it done. What I have is...

$$-1000e^{-500t}-2000e^{-1000t}=0$$

My first inclination is to use ln, but ln(0) is undefined. Any tips?

 

[sjb-2812]

Are you sure that equation is right? Looks to me like you have -(something which is always positive) - (something else that is always positive) = 0, but the LHS will always be negative, and never zero, unless t is infinite?

 

[CRGreathouse]

That would seem to indicate that the power has no maximum; perhaps it increases without bound.

 

[willem2]

I'm trying to calculate maximum power. Once I computed power as a function of time I took its derivative and set it equal to zero. Now it need to solve for time and I can't seem to get it done. What I have is...

$$-1000e^{-500t}-2000e^{-1000t}=0$$

My first inclination is to use ln, but ln(0) is undefined. Any tips?

As it is now, there's no solution. You probably made a sign error while computing the derivative.
If that is indeed the case you can try substituting $e^{-500t} = u$

 

[CentreShifter]

Thanks to all who responded.

I believe I was doing my computations incorrectly. I am given voltage:

$$100e^{-500t}V$$

And current:

$$20-20e^{-500t}mA$$

Power being the product:

$$2e^{-1000t}(e^{500t}-1)W$$

And derivative:

$$-1000e^{-1000t}(e^{500t}-2)$$

To find out when power is maximized:

$$-1000e^{-1000t}(e^{500t}-2)=0$$

$$2000e^{-1000t}=1000e^{-500t}$$

$$ln2=500t$$

Putting t at about .0013863 s.

 

[Redbelly98]

I can't find any error so far.

Just out of curiousity, what is the purpose of the circuit?

 

[CentreShifter]

It isn't real. This was one part of a practice problem I was solving as to lead me up to DEs, which I need to reacquaint myself with.

 

[Redbelly98]

Oh. Questions like that should get posted in the Homework & Coursework Questions area, even if it's for personal study and not an actual homework assignment.

I have moved this thread.

 

[CentreShifter]

Cool. Thanks for the info.