Question

solve for t
t+1=x(2^t(t-1)t)

Answer 1

\[t+1=x(2^t(t-1)t) \]
seems like it?

Answer 2

θ(1+x)^-(1+θ),x>1

Answer 3

can someone intergrate from 1to infinity

Answer 4

t+1=x(2^t(t-1)t)
t =x(2^(t(t)-t)(t))
is it like ths

Answer 5

\[t+1=x(2t ^{2}(t-1))\]

Answer 6

it's a cubic equation in t.

Answer 7

yes but i don't get t

Answer 8

there is a brute method
http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

Answer 9

wen i simplify i got
\[\frac{ (t+1) }{ t-1 }=x(2t ^{2})\]

Answer 10

this doesn't look easy
http://www.wolframalpha.com/input/?i=solve++t%2B1%3Dx%28t^2%28t-1%29%29+for+t

Answer 11

\[ 2x t^3 - 2x t^2 - t - 1 = 0\]
directly use this http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots to get the roots of cubic equation.

or, make this type of substitution http://en.wikipedia.org/wiki/Cubic_function#Reduction_to_a_depressed_cubic to get cubic in depressed form and use Cardano's method to get the roots http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

Answer 12

this might be easy for particular value of x.

Answer 13

I suppose there is something wrong somewhere.
I'm thinking the t + 1 is really t - 1. (that is my thinking)