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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.

steve
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    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 $t^{1.44} = 1.44^t$". – Arthur Sep 11 '14 at 15:08
  • 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. – steve Sep 11 '14 at 15:14
  • There are no negative solutions. $t^{1.44}$ coes into complex when $t$ goes negative. For absolute values, though, the solution is there. – Arthur Sep 11 '14 at 15:16
  • 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. – steve Sep 11 '14 at 15:18
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    Your calculator plots the absolute value, I'm sure. Check this WA link. – Arthur Sep 11 '14 at 15:19
  • That's exactly what happened. Thanks so much for clearing that up for me! – steve Sep 11 '14 at 15:19
  • @Arthur: $t^x$ is a well-defined real number for $t < 0$ if $x$ can be expressed as a fraction $p/q$ with $q$ odd, e.g. $x = 1.44 = 36/25$. – TonyK Sep 12 '14 at 18:14
  • @TonyK which is almost nowhere, and almost certainly not at this $-0.81377\ldots$ point. – Arthur Sep 12 '14 at 21:18
  • @Arthur: $1.44^t$ is defined because $1.44 > 0$, and $t^{1.44}$ is defined because $1.44 = 36/25$. – TonyK Sep 12 '14 at 21:20
  • If a and b are positive real, there will be at least 2 solutions for a^b = b^a. Right? – Narasimham Sep 16 '14 at 19:28
  • There are different conventions for $t^x$ when $t<0$. One that works quite generally is using complex numbers and picking a principal value using some convention. Another, conflicting, one will work only in the cases @TonyK mentions. In this example, the latter becomes equivalent to $$t^{1{.}44} = (t^{36})^{1/25}$$ where $t^{36}$ is well-defined because the exponent is an integer, and also non-negative and real (for $t$ real), and so the next exponentiation is a positive number raised to a positive number. – Jeppe Stig Nielsen Jul 25 '18 at 20:54

3 Answers3

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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.

flawr
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  • Did you see Arthur's comment about the second solution near t= 8.040854? – mike Sep 11 '14 at 15:39
  • +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) – almagest Sep 12 '14 at 16:21
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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:

$$\frac{\log(1.44)}{1.44}=0.253224=\frac{\log(8.40854)}{8.40854}\tag{12}$$

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

mike
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  • 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. – flawr Sep 11 '14 at 15:13
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    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. – mike Sep 11 '14 at 15:16
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    This function is also known as the Lambert W Function – Alice Ryhl Sep 11 '14 at 15:20
  • @mike then please explain how you can analytically solve it. – flawr Sep 11 '14 at 15:22
  • see my figure. it is visible there – mike Sep 11 '14 at 15:23
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    @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? – Jam Sep 11 '14 at 15:24
  • The purple curve is $t(x):=-\frac{W(x)}{x}$ – mike Sep 11 '14 at 15:27
  • I missed a minus sign in my previous comment. The purple curve is $t(x):=-\frac{W(-x)}{x}$ – mike Sep 11 '14 at 15:36
  • @EulCan How is the above an analytic solution? – almagest Sep 12 '14 at 08:37
  • @almagest It gives the solution in terms of well-defined, known functions (with zero error) as opposed to a decimal with some error. – Jam Sep 12 '14 at 15:41
  • @EulCan Really? So what is the expression in terms of such functions? – almagest Sep 12 '14 at 15:48
  • @algamest It's the solution mike gave in his answer; $t=\frac{-1.44\mathrm{W}\left(-\frac{\ln(1.44)}{1.44}\right)}{\ln(1.44)}$. – Jam Sep 12 '14 at 15:54
  • @mike Important update: I've realised (after finding how to use W in wolfram.alpha) that your solution simplifies to $1.44$ (which is why it works); it isn't the solution around 8.0409. Check this link: http://bit.ly/1pd0Q5A – Jam Sep 12 '14 at 16:09
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    @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"! – almagest Sep 12 '14 at 16:10
  • @almagest You're correct that it simplifies to $1.44$; I only just checked it now as wolfram.alpha didn't process it properly (it would've been helpful if other people would've checked the solution before upvoting though). I don't know what mike meant. You'll have to ask him yourself. – Jam Sep 12 '14 at 16:12
  • I did, and got a totally unsatisfactory answer! See comments below my answer, which was downvoted because I said I could not see an analytical solution! – almagest Sep 12 '14 at 16:18
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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.

almagest
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  • please see my solution – mike Sep 11 '14 at 15:36
  • @mike see comment added below your solution. – almagest Sep 11 '14 at 19:31
  • @mike and now additional material above. – almagest Sep 11 '14 at 21:05
  • 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. – mike Sep 12 '14 at 01:17
  • @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. – almagest Sep 12 '14 at 05:48
  • It has nothing to do with LambertW function. I just figured out that it is due to the following strange identity: $\frac{\log(1.44)}{1.44}=0.253224=\frac{\log(8.40854)}{8.40854}$. – mike Sep 12 '14 at 05:54
  • $\frac{\log(1)}{1}=0$,$\frac{\log(\infty)}{\infty}=0$. So $\frac{\log(x)}{x}$ is unimodular, takes max value at $x=e$. – mike Sep 12 '14 at 06:00
  • @mike I have to say that I find your solution totally confusing. You start by mentioning $W(x)$. You appear to be implying that there is an analytic solution in terms of $W(x)$ - both I and EulCan interpreted it that way, but you are now confirming that was not what you meant. Also this comment thread started because you asked me to see your solution. Why? What does it add? – almagest Sep 12 '14 at 08:34
  • For the record I did not down-vote your solution. – mike Sep 12 '14 at 17:48