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Please give me a hint in order to solve in t, the following equation:

$$t^2e^{a^2\!/t} + a^2e^t-12at=0$$

where $a=\ln(4)$.

Bernard
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riciu
  • 11
  • I can suggest two improvements to your question. First, please typeset your math using mathjax, otherwise no-one will want to read your question. Second, explain what you've tried, where you got stuck, and in general avoid "no-clue" questions. – Lee Mosher Jul 12 '19 at 14:13
  • are you looking for "exact" or approximate solution ? – G Cab Jul 12 '19 at 14:30
  • First of all I need a hint... Can't exponentiate it, can't use W lambert. In fact I can, but, this doesn't do any good, because I obtain some ugly expression. If I divide by e^((a^2)/t)) or by e^t, is no use either. Of course I can change the terms and so on. Of, and I can't raise to the power of t, or (a^2)/t), either, because again I obtain something ugly. As I said, I don't need to be solve by you, but only a hint. And just for the record, t is a substitution for log(x), from another equation. I look for an exact solution, because from there I must put it in another equation. – riciu Jul 12 '19 at 14:31
  • Newtons method – user90369 Jul 12 '19 at 15:01

1 Answers1

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Hint: plug in $t:=a\cdot x$ for a new unknown $x$. This simplifies the equation a bit and (provided I didn't make a computation error) if you plug in $a=\ln(4)$ and then try some small integers for $x$ you will find a solution.

Edit: You already found $x^2*e^{a/x}+e^{ax}-12*x=0$. Now plug in $a=\ln(4)$ gives $$x^24^{1/x} + 4^x = 12x$$ Pluggin in $x=2$ gives a solution.

Note that this uses being lucky with your specific equation. If you replace the $12$ by a $13$ I don't think there is any way to solve this algebraically.

quarague
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  • I will let you know about what I did, probably tomorrow. Thanks. – riciu Jul 12 '19 at 14:41
  • This don't work either... (a^2)(x^2)(e^((a^2)/(ax)))+(a^2)e^(ax)-12(a^2)x=0 Which mean ((x^2)(e^((a/x)))+e^(ax)-12x=0, and *** this close to my initial eqation to solve, which I didn't mention here yet... in fact is the same pattern. – riciu Jul 13 '19 at 09:59
  • 0.5 is also a solution. But I don't see how you solved it... I design myself the equation so I know that 0.5 and 2 are solution. I could easily solve it by plotting. But, I wanted to know how to solve it, and I didn't find a way of solving, except plot and guesses... The last doesn't apply because I know the answers... So, if there are no ways of solving other the plotting or / and "try and guess", bisection method and other "not-exacted" ways ... please tell me so. Thanks. – riciu Jul 19 '19 at 10:50
  • I found the solution $x=2$ by guessing. I don't think there is a way to solve this kind of equation algebraically. – quarague Jul 19 '19 at 10:52
  • Ok. Thanks. I only played, and it was only for (my) fun. I have designed a lot of similar kind of equation, but only in some special cases I could find a solution without plots and try and guesses. – riciu Jul 19 '19 at 11:01