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- Thread starter jeffdj
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In summary, the conversation discusses solving for x in the equation f(x) = x*log(x) = c, where c is a constant. One solution is x = 10 and c = 10, but there is no closed form solution and the use of the Lambert W function is considered a "trick" for solving. The conversation also touches on the question of whether there is a way to prove that there is no closed form solution, and the use of calculus and other tools beyond algebra for solving the problem.

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because both of them are the same f(x) then: x*logx=c

logx^x=c

10^c=x^x

now one solution of this is x=10 then also c=10.

was this the solution you were after?

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The exact problem that I seek a solution to is:

10^6 = x*(log x/log 2)

I'm attempting to solve for x. I suppose I can use Maple when I get access to it on Tuesday, but I would like to understand the math behind the solution. Thanks for your help!

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cheers,

phoenix

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x= W(c) where W is the Lambert W function.

Other than that, there is no "closed form" solution.

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x= W(c) where W is the Lambert W function.

Other than that, there is no "closed form" solution."

just curious: is there a way to prove that there is no closed form solution? i believe that as well.

i do consider the lambert w function a trick, but not a bad one. whenever functions can't be inverted using "normal" functions, new ones get invented. the log and square root are typical examples of this as being inverses of functions that can't "normally" be inverted. so it looks like you will have to go to maple on tuesday and use a root finder to solve it numerically.

amazing how simple algebra questions can lead to deep questions, huh?

may your journey be graceful,

phoenix

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Find all numbers x such that:

x+3

To rewrite,

3

All negative numbers will work since a negative value for x will make 3^x smaller than 3 and 4-x greater than 4.

In fact, it seems that all numbers less than one work.

3^1=4-1

These are both continuous fxns. And d3^x/dx=ln3*3^x=ln27>0 at x=1. d(4-x)/dx=-1 at x=1.

So they intersect at x=1 and from there the left hand side increases while the right decreases. Imagining the graphs of these two, it's easy to see that x<1. However, this appears in a text BEFORE calculus is introduced. So how to solve with algebra?

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how did you encounter this strange question?

i think you're going to have to permit yourself to use calculus and other tools beyond calculus for this one or at least be open to the possibility that algebra is not a sharp enough instrument to tackle this question with. sorry if that's sounding like bad news but hey, algebra has its limits (no pun intended).

cheers,

phoenix

i think you're going to have to permit yourself to use calculus and other tools beyond calculus for this one or at least be open to the possibility that algebra is not a sharp enough instrument to tackle this question with. sorry if that's sounding like bad news but hey, algebra has its limits (no pun intended).

cheers,

phoenix

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Oh well, it's no matter.

Thanks.

The "Solve X * log(x) Trick" is a mathematical technique used to solve equations that involve both a variable, represented by X, and the logarithm function, represented by log(x). This trick allows for the simplification and solution of complicated equations involving these two components.

The trick involves taking the logarithm of both sides of the equation, which allows for the variable to be isolated and solved for. By using properties of logarithms, the equation can be simplified and the variable can be found.

This trick is most useful when solving equations that involve both a variable and a logarithm, and where the variable is present as both a base and an exponent. It is also helpful for simplifying and solving equations with complicated expressions involving logarithms.

Yes, when applied correctly, the "Solve X * log(x) Trick" will always provide an accurate solution to the equation. However, it is important to double check your work and ensure that you have not made any mistakes during the simplification process.

While this trick is a useful tool for solving equations involving logarithms, it may not be applicable to all equations. In some cases, it may not be the most efficient method for finding a solution. It is important to consider other techniques and approaches when solving equations.

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