Solve for t: If x=(e^t) - (e^-t)?

Question

I don't remember how to apply natural LOG rules.

Thank you Pascal, that makes a lot of sense. But, it makes things a little more complicated for the entire problem. I guess I should have mentioned the rest. I'm trying to find arc length where

x=(e^t) + (e^-t)

y=5-2t where 0≤t≤3

I the answer, wich I think is (e^3) - (1/e^3)

But I don't know how to graph the actual arc. If I use the quadratic equation, won't I get 2 different t values?

I think i opened a can of worms here

Answer 1

Here is a way of solving for t. The series expansion for e^t is

e^t = sum(i=0, infinity, (t^i)/i!) where i! is factorial(i).

This series converges over all the complex plane.

Then e^t - e^(-t) would be

sum(i=0, infinity, (t^i)/i!) - sum(i=0, infinity, ((-t)^i)/i!)

Notice that the even terms cancel and the odd terms are doubled (add each even term to itself). Then

e^t - e^(-t) = 2 (sum(i=0, infinity, t^(2i+1)/(2i+1)!)

But the term in parenthesis is the series for sinh(t), which also converges over all the complex plane. The original equation thus becomes

x = 2 sinh(t)

or sinh(t) = (x/2). If we take arcsinh() on both sides we get
the family of solutions (there is more than 1):

t = (-1)^k arcsinh(x/2) + k pi i, for all integers k

where pi = 3.14159... and i = sqrt(-1)

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I read your comments. Well, not quite a can of worms. Your answer for the arc length is correct. Notice that the function I gave you above (lets ignore all complex values and choose only the positive solution) is

t = arcsinh(x/2)

Since y = 5 - 2t then

y(x) = 5 - 2 arcsinh(x/2)

and you can easily graph this using tables for arcsinh.

In all cases you will have to express t as a function of x, and since y is a function of t, then you can express easily y as a function of x and use that to graph the arc portion in [0,3]

Answer 2

First, multiply by e^t:

xe^t = e^(2t) -1

This is now a quadratic equation in e^t, so placing it in standard form:

e^(2t)-xe^t-1=0

And solving using the quadratic formula:

e^t = (x±√(x²+4))/2

Now simply take the logarithm:

t = ln ((x±√(x²+4))/2)

And we are done. No manipulation of logarithms was necessary (aside from taking a single log at the end), the key here was to recognize that this was primarily a quadratic equation, rather than a logarithmic one.

Answer 3

t = Log(1/2 ± √(x²+4))

The express e^t - e^-t is actually the same as 2Sinh(t), or the hyperbolic Sine. See wiki.

Addendum: Pascal gave a good explanation of the solution before I could get it down here. Give him the 10 points.

Answer 4

This will actually yield an expression in quadratic form.
Let u be e^t.
x = u - 1/u
==> u^2 -xu - 1 = 0
by quadratic formula: u = [ x {+-}sqrt(x^2+4) ]/2
since u is exponential, it is positive.
u = [x+sqrt(x^2+4]/2=e^t
then t = ln u = ln([x+sqrt(x^2+4]/2)

Answer 5

I wish I knew math, it has never made any sense to me, but I love watching others solve it.

Answer 6

You can use a trig identity, sinh t = (e^t - e^-t)/2,
thus 2x = sinh t, and t = inverse sinh (2x)

Answer 7

t=Log(x+((x^2)+4)^1/2)

Answer 8

x=(e^t) - (e^-t)
x/2 = (e^t - e^-t)/2
x/2 = sinh t
t = sinh^-1 (x/2)