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Throughout my high school career I was always told that an equation of this sort ( $e^x = 6x$ for example) couldn't be solved algebraically. However I feel that there may be a way (and you may be out there saying "of course there is a way") I know that it can be solved graphically, but is there any other way(s) to solve this equation: $$e^x = 6x$$ **Without graphing or using a equation solver **

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  • $\begingroup$ What you were told is, roughly speaking, correct. The closest you can get to an 'algebraic' solution to that equation would require the use of some Special Function in order to isolate the $x$ you intend to solve for (because it occurs in an exponent in one place, and not in an exponent elsewhere). $\endgroup$ – Justin Benfield Apr 10 '16 at 4:11
  • $\begingroup$ No, this is a transcendental equation. It can be solved approximately through numerical methods. $\endgroup$ – Jared Apr 10 '16 at 4:11
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    $\begingroup$ @WaveX Have you tried the newton-raphson method ? 2 can be choosen as the initial value. $\endgroup$ – callculus Apr 10 '16 at 4:16
  • $\begingroup$ @Justin Benfield I assume you are referring to the Lambert W function? Granted I haven't learned enough about it yet to know how it works. $\endgroup$ – WaveX Apr 10 '16 at 4:16
  • $\begingroup$ If your going to accept lambert w though, then the trick is to let $-u=x$ and continue from there. $\endgroup$ – Ahmed S. Attaalla Apr 10 '16 at 20:07
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The term "solved algebraically" is a bit tricky, because you have to specify what functions are permitted in the "algebraic" solution. For example, if you don't allow the square root function, then even the equation $x^2 = 2$ can't be solved algebraically.

So, when people discuss equation solving, they usually talk about solutions expressed in terms of "radicals", which includes square roots, cube roots, and so on. And the use of simple trig functions (like sine, cosine, tangent) is usually permitted, too.

With these sorts of restrictions, equations like $e^x = 6x$ can not be solved "algebraically" -- there is no formula for the solution. I believe there are proofs of this; certainly one can prove that polynomial equations of degree higher than four can't be solved by radicals. But the proofs would be difficult to understand unless you have a good mathematical background.

In practice, if you just want to calculate a solution, the existence (or not) of an algebraic method doesn't matter very much. You can always use numerical methods, like the Newton-Raphson algorithm, to find roots, instead. And, in fact, even when formulae do exist, people still use numerical methods, sometimes. For example, there are formulae giving the roots of cubic and quartic equations. But they are complex, and it's difficult to go through all the computations without numerical errors building up. So, numerical methods might give more accurate answers.

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  • $\begingroup$ I particularly like this answer because we generally think of "solvable algebraically" to mean through elementary operations like addition, subtraction, multiplication, and division. But division itself can be very "difficult" to perform...usually we gloss over this with the rational numbers, e.g. $1\div 3 = \frac{1}{3}$. If you think about how to get the decimal value of $1\div 3$ it's not all that dissimilar to how you would find the solution to $e^x = 6x$ via an iterative method such as Newton's method. $\endgroup$ – Jared Apr 10 '16 at 5:59
  • $\begingroup$ @Jared -- Yep, as soon as you start using a computer, almost everything is an approximation. The approximations used in doing division or computing square roots are somewhat hidden, so it's easy to overlook them, as you say. The approximations involved in Newton-Raphson methods are more visible, but they are not necessarily less accurate or more dangerous. $\endgroup$ – bubba Apr 10 '16 at 9:03
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You can use the Lambert-$W$ function to write an "algebraic" solution, or at least an 'explicit' solution. As pointed out in the comments, let $-u=x$: $$e^{-u}=-6u\Rightarrow\frac1{e^u}=-6u$$ $$\frac1{ue^u}=-6\Rightarrow ue^u=-\frac16$$ From the definition of the special function, $ze^z=k$ can be written as $z=W(k)$, and thus: $$u=W\Big(-\frac16\Big)$$ $$\bbox[10px, border:1px solid black]{\therefore x=-W\Big(-\frac16\Big)}$$

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    $\begingroup$ Thanks for the step by step answer. There's a small typo in your second line though, "$ue^e$" I'm guessing should be $ue^u$ :) $\endgroup$ – WaveX Apr 2 '18 at 18:57
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I am a fan of fixed point iteration because I find it easy to set up. You want to work your equation into the form $x=f(x)$ where the derivative of $f(x)$ at the root is small and certainly less than $1$ in absolute value. Then pick a reasonable starting value for $x_0$ and iterate $x_{i+1}=f(x_i)$. As logs are slowly varying I would write this as $$x=\log (6x)=\log (6) + \log (x)$$ A starting value of $x_0=\log(6)$ looks reasonable. After a couple dozen iterations in a spreadsheet (love copy-down) it has converged to about $2.833148$

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