Solving $(1.44)^t=t^{1.44}$ I've been trying to solve the equation  $(1.44)^t=t^{1.44}$, but other than the obvious solution ($t=1.44$) I haven't had much luck manipulating this into something useful. By taking the log of both sides I'm able to get $\dfrac{\ln t}{t}=\dfrac{\ln 1.44}{1.44}$, but then I'm left with essentially the same problem--my variables are in two different "places" and I can't figure out how to combine them. I can also use exponents to rewrite this as $t^{\frac{1}{t}}=e^{\frac{\ln 1.44}{1.44}}$ or $t^{t^{-1}}=e^{\frac{\ln 1.44}{1.44}}$.
I also tried rewriting the equation as $(1.44)^t-t^{1.44}=0$ and factoring out $1.44-t$, but it quickly turned into a complete mess. Any thoughts on how to approach this? I know two other solutions exist by looking at the graphs of the two functions.
 A: One 'obvious' solution is of course $t=1.44$ for any other solutions you need to use a numerical method, because as you realized too, you cannot solve the equation for $t$ analitically.
A: Let $W(z)$ be the solution of $z=We^W$, then the solution to the original equation is:
$$t(a)=-\frac{W(-(\frac{\log(a)}{a}))}{\frac{\log(a)}{a}},a=1.44 \tag{2}$$
One trivial numerical solutionis $t(1.44)=1.44$.
$W(z)$ is the Lambert W Function. It is a build-in function in Mathematica(7.0) called ProductLog.

Here is a plot of $t(a)$ vs. $a$ (purple). It is mentioned that there is another solution near t=8.040854. We can see it from the figure. The horizontal blue line is $t=1.44$
EDIT (2014-09-12).
There seemed to be some confusion about how many solutions the original equation (shown below) has:
$$(1.44)^t=t^{1.44} \tag{10}$$
Taking the $\log$ on both sides we can rewrite it as:
$$\frac{\log(1.44)}{1.44}=\frac{\log(t)}{t} \tag{11}$$
The first solution to (10) is obvious: $t=1.44$
Numerical result (by @Arthur) also showed that there exists a second solution: $t=8.40854$.
This is because:
$$\frac{\log(1.44)}{1.44}=0.253224=\frac{\log(8.40854)}{8.40854}\tag{12}$$
The question then is: are there any other solutions to (10)?
I do not know how Arthur got the second numerical solution. By expressing the solution to (10) as in (2), we can just plot t(a) vs. a to see if there are any more unexpected solutions.
For example, we can say that the following equation (13) has no real solution for t:
$$b^t=t^b\qquad b>e=2.71828...\tag{13}$$
All of this can only be achieved after we express the solution to (10) as in (2) and we carry out the numerical experiments based on the knowledge of Lambert W Function.
Now we can ask the following question. 
$$\exp\left(\frac{\log(1.44)}{1.44}\right)=\exp\left(\frac{\log(t)}{t}\right)\tag{14}$$
How many solution of $t$ exist for equation (14)?
It is now obvious to us that there are two solutions: $t=1.44$ and $t=8.40854$.  
This is why I made some comments before saying that the second solution to (10) has nothing to do with Lambert W Function. It also showed up in (14) with $\exp$ function.
Hope my explanation helps-
mike
A: 
Here is a plot. Showing the two curves (there is no real value for $t^1.44$ when $t<0$, so that curve is only plotted for positive $t$.

You can also take logs. Here is the plot for that. 
Unfortunately, I cannot think of any way of getting the exact value of the larger solution in terms of elementary functions (or even, at the moment, "special" functions).
[Added later] Just to be clear, Mike's solution (at 0839Z 12 Sep 14) is not an analytic solution, although there may be one if someone could just think of it.
