How Do I Solve for X in f(x) = c Using Logarithms?

ill have a try

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?

Well I had kind of made it that far. I felt like I was going in circles.

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!

proving this claim may be difficult, but i believe it is impossible to algebraically solve for x in ways that involve normal functions.

cheers,
phoenix

You might consider it a trick. The "Lambert W function" is specifically defined (by Euler- he named it for the same man the "Lambert quadrilateral" is named for) as the inverse of the function f(x)= x ln(x). The solution to the equation xln(x)= c is
x= W(c) where W is the Lambert W function.

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

"You might consider it a trick. The "Lambert W function" is specifically defined (by Euler- he named it for the same man the "Lambert quadrilateral" is named for) as the inverse of the function f(x)= x ln(x). The solution to the equation xln(x)= c is
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

I have a similar question.

Find all numbers x such that:
x+3^x<4

To rewrite,
3^x<4-x
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?

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

That's what I suspected. It comes from a Calc book that's meant to give a rigorous formulation of the subject starting with properties of numbers and working up to limits, derivatives and all that. It's a problem in the back of the first chapter, which is called "Properties of Numbers." I don't know how the author expects me to solve it w/o calc. Maybe finding when they are equal and using test points to either side. But that requires knowledge of the concept of continuity, which hasn't been discussed.
Oh well, it's no matter.
Thanks.