what is the answer to t^-t=8/e ?

Question

tried to solve the following:
f(x)=2log x (or log x/log2)
F(t)-F(1)=3
what is t?

I came as far as this:

F(x)= 1/ln 2 * (x ln x - x)

so 1/ln 2 * (t ln t - t) + 1/ln 2 = 3

so t ln t - t = 3* (ln 2) - 1

so ln (t^-t)= ln 8 - ln e

so t^-t = 8/ e

and that's where i'm stuck, so i'm asking how can i get any further in finding an exact answer to what "t" is? and am I on the right track or have I made a flaw already?

Answer 1

Stop, stop, stop. If it was f(x)=2logx, f(x) cannot be = to (1/log2)*logx, as 1/log2 is not = 2.
Don’t hurry, put your question correctly with necessary details and don’t save on ()-s in formulas.

Answer 2

This problem is not solvable using ordinary functions. Therefore the lanbert W function is needed in order to get a solution.
You have t^(-t)=8/e. Take the inverse of both sides.
t^t=e/8

e^(t*Ln(t))
Take the logarithm of both sides
t ln(t)=Ln(e)-Ln(8)=1-Ln(8)
Now let t=e^v. You get
v*e^v=1-Ln(8)

The lambert w function is defined as follows.
W(x*e^x)=x.

Take the lambert-W function of both sides.
W(v*e^v)=W(1-Ln(8))

v=W(1-Ln(8)

e^v=e^W(1-Ln(8)

t=e^v therefore

t=e^W(1-Ln(8))

Answer 3

no your analysis doesnt seem to be right.
F(x) = {xlnx-x}/ln2
and u are simply replacing x by t:
F(t) = {tlnt-t}/ln2 ...........(1)
and that for t = 1, automatically:
F(1) = -1/ln2
and that F(t) = 3+F(1) = 3-1/ln2 = {3ln2 -1}/ln2 .......(2)
comparing (1) and (2)
tlnt-t = 3ln2 -1
= 2ln2 +ln2 -2+1
= 2ln2-2 +ln2+1
=2ln2 -2 +ln(2/e)
now since ln(2/e)~0 as 2/e ~1, hence t = 2 approx.

Answer 4

t^-t = 8/e
t^-t = 8/2.7182818 = 2.943
take log of both sides:
-t log t = log 2.94
-t log t = .468
if t is positive, the result will be negative, and t cannot be negative, so there is no solution.

Answer 5

question not clear