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I'd like to know if I can solve the following equation without calculator:

$(0.4)^t=5t$

I don't think it's possible, cause I always get stuck on formulas of the form $e^t=t$ or $t=\ln t$

I've also put the equation into wolframalpha, which was of no use to me unfortunately.

I'm not interested in the answer containing a W-function. Just want to know whether I can find the real solution or not!

Thanks!

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Here is a more general case. –  Mhenni Benghorbal Jan 17 '13 at 5:15

3 Answers 3

No, it is not possible to solve in terms of elementary functions. That is why W|A returns an answer using the W function.

When you find the formula boils down to $t=\ln t$, you've come across a form of Lambert's Transcendental Equation:

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Did I really just get an upvote in 16 seconds? That's a little weird. :) (but in a good way) –  anorton Jan 17 '13 at 2:17
    
Okay thank you for your quick answer! :) –  Quickquestion Jan 17 '13 at 2:17
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@Quickquestion My pleasure. Since you're somewhat new to Math.SE, I will remind you that you can accept answers that, in your opinion, answer your question the best. That is done by clicking the green checkmark on the left of the answer you like. This makes the accepted answer stay on top of all the other answers, provides a reputation bonus to the answerer, and increases your own reputation by 2pts. –  anorton Jan 17 '13 at 2:23

You will need a numerical method or some not-so-obvious special function to do this.

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There is a solution $t$, since $(0.4)^t\gt 5t$ at $t=0$, and $(0.4)^t\lt 5t$ at $t=1$. It is easy to see that there cannot be a solution outside the interval $(0,1)$.

There is no rational solution. For suppose that $t=\frac{p}{q}$ is a solution, where $p$ and $q$ are relatively prime integers, neither equal to $0$. Since $0\lt t\lt 1$, we have $p\lt q$. Then from $$\left(\frac{2}{5}\right)^{p/q}=5\frac{p}{q}$$ we obtain $$2^pq^q=5^{p+q}p^q.$$ Thus $p$ is a power of $2$. This is impossible, since then $2^p \lt p^q$.

By using the Gelfond-Schneider theorem, we can now prove that $t$ cannot be an irrational algebraic number. For if $t$ is irrational algebraic, then $(0.4)^t$ is transcendental, but $5t$ is not.

So $t$ is transcendental. We have not ruled out the possibility that $t$ is a simple combination of "simple" transcendentals, such as $\log 2$, $\log 3$, \sin 1$, and so on.

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