exponential math question?

Question

i really need help. cracked my brains over this for like 30 mins already.

t - 2e^(-0.5t) = 2

how do you solve for t?

Answer 1

Is this a problem in Complex Analysis?

There are no "real" solutions for t, but there may be complex values (of the form a + bi) for t that would work.

You can prove that real values don't work by creating a graph (with a graphing calculator or using Excel, for example). the expression on the left does not equal 2 for any real values of t.

Answer 2

Hi,
I have kinda forgotten asto how to complee this.

Given,
t-2e^(-0.5t) = 2
=> e^(-0.5t)-0.5t+1 = 0 (take 2 to other side, muitlply by -0.5)
=>e^X-X = -1 where X=-0.5t
Taking log to base e ie., ln
=> X-ln(X) = -ln(1)
=> X = ln(X) {ln(1) = 0)}

I have come till here. Its all your's

Sorry, cant complete it.

Here are a few links that didnt quitehelp me . May be they will help you.



You can try using the complex equations with X=0.5t instead of (-0.5t)

then we have e^(it)=-(1+t)
and
e^(it)= cost + isint

=> cost + isint = -(1-t)

Peace out.

Answer 3

Plotting the graphs y = t-2e^(-t/2) and y = 2 shows one intersection which implies there is one real solution for t.

we will enter the realms of complex analysis anyway.

Let z=t/2 so the equation becomes

2*z - 2e^(-z) = 2 or

z - e^(-z) -1 =0

Now if z is complex then we can write z = x+iy where x and y are real.
so

x+iy - e^(-x-iy) - 1=0

using e^(-iy) = cos(y)-isin(y) we get

(x-1 - cos(y)*e^(-x)) +i(y + sin(y)*e^(-x)) = 0

This implies that both the real and imaginary parts must be zero giving two simultaneous equations for x and y:

x-1 - cos(y)*e^(-x) = 0 and
y + sin(y)*e^(-x) = 0

solving these will get you x and y. Then z=x+iy and z=t/2

so t = 2x+2iy

Answer 4

Its not possible to rearrange this equation to make t the subject.
You have to solve iteratively - computers are so quick nowadays it's no big deal.
t=2.556929 (approx.)

30 mins is not long to spend on a problem!

Answer 5

by taking 2+0.5+2e^

Answer 6

get someone who knows more about it than I do.