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.

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    $\begingroup$ There is one other solution, at about $8.040854$, but there's no nice way of describing the exact value, other than "The non-trivial solution to $t^{1.44} = 1.44^t$". $\endgroup$ – Arthur Sep 11 '14 at 15:08
  • $\begingroup$ I think there's another solution at -.81377 as well. It's nice to know that I'm not just missing something, and you actually need numerical methods to get this answer. $\endgroup$ – steve Sep 11 '14 at 15:14
  • $\begingroup$ There are no negative solutions. $t^{1.44}$ coes into complex when $t$ goes negative. For absolute values, though, the solution is there. $\endgroup$ – Arthur Sep 11 '14 at 15:16
  • $\begingroup$ Weird. My calculator is able to graph t^{1.44} and it looks like it's close to mirrored square root oriented about the $y$-axis. It must be interpreting it in a weird way. $\endgroup$ – steve Sep 11 '14 at 15:18
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    $\begingroup$ Your calculator plots the absolute value, I'm sure. Check this WA link. $\endgroup$ – Arthur Sep 11 '14 at 15:19

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.

  • $\begingroup$ Did you see Arthur's comment about the second solution near t= 8.040854? $\endgroup$ – mike Sep 11 '14 at 15:39
  • $\begingroup$ +1 Mainly, I admit because I am so irritated with Mike's answer (which technically does not claim an analytical solution, but most people upvoted it because they thought it provided one) $\endgroup$ – almagest Sep 12 '14 at 16:21

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.

enter image description here

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:


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.


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

  • $\begingroup$ I do not think that this helps since you have to solve $z=we^w$ numerically. This way you could solve the original equation numerically. $\endgroup$ – flawr Sep 11 '14 at 15:13
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    $\begingroup$ I was checking if there are solutions other than the obvious one (t=1.44). I do not have to solve $z=we^w$ numerically. $\endgroup$ – mike Sep 11 '14 at 15:16
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    $\begingroup$ This function is also known as the Lambert W Function $\endgroup$ – Alice Ryhl Sep 11 '14 at 15:20
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    $\begingroup$ @flawr Mike has given an analytic solution; $W(x)$ is a known function in the same way that $\sin(1)$ or $e^2$ are known constants even though they're defined by their functions. Eg. $W(1)$ is a closed form & not an approximation . Mike, sorry but what's the purple curve in the graph? $\endgroup$ – Jam Sep 11 '14 at 15:24
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    $\begingroup$ @EulCan No. That evaluates to 1.44. Check it on Mathematica. Anyway seem Mike's comment below my solution. I asked "But what is 8.40843 in terms of the Lambert function? "; he replied "It has nothing to do with LambertW function"! $\endgroup$ – almagest Sep 12 '14 at 16:10

enter image description here

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 $t$.

enter image description here

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.

  • $\begingroup$ please see my solution $\endgroup$ – mike Sep 11 '14 at 15:36
  • $\begingroup$ @mike see comment added below your solution. $\endgroup$ – almagest Sep 11 '14 at 19:31
  • $\begingroup$ @mike and now additional material above. $\endgroup$ – almagest Sep 11 '14 at 21:05
  • $\begingroup$ You missed a minus sign in your plot function. In Mathematica define 't[a_]:=LambertW[-Log[a]/a]/(-Log[a]/a)'. If you carry out the following plot: 'Plot[{144/100,t[a]},{a,1.2,10}]', then you will see the plot like shown in my answer. This means that t[a]==1.44 has two solutions: one is t[1.44]==1.44; the other is t[8.40854]=1.44. $\endgroup$ – mike Sep 12 '14 at 1:17
  • $\begingroup$ @Mike Thanks for the correction on the plot. I will fix later this morning. But what is 8.40843 in terms of the Lambert function? Must rush now - running late. $\endgroup$ – almagest Sep 12 '14 at 5:48

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