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How do you solve equations like this Veeky Forums
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take the xth root
It's a troll. There's only two solutions.
>posts a link showing there are three solutions
0=sqrt(2)*2^x - x
Numerically.
bumping I want to know this to
if you have quadratics possible then graph it this is high school math user
"graph it and look" is a brainlet-tier solution, not at all generalizable.
This class of problem is deceptively simple user, a quick investigation reveals that solving [math]x^k = k^x[/math] involves the Lambert W function which is non-elementary, that is it cannot be described in terms of a finite number of elementary operations.
bump
2,4, use Newton's method starting at -1 with f(x)=x^2-2^x. You can get as close as you want.
Dropped math after high school. What's a Lambert W function?
>Scrolls to integer solutions
>Doesn't realize one of the solutions is a non-integer.
-7.666 repeating,
2,
and 4
x^2 = 2^x
log2(x^2) = log2(2^x)
x * log2(2) = log2(2^x)
x = log2(2^x)
logx(x) = x * logx(log2(2^x))
1 = ?
I give up. Is there a way to do this automatically or do you literally have to do that complete the square bullshit?
Nigga what?
It's x=1,2,4
1^2 = 1
2^1 = 2
But one of them is negative number.
Oh sorry I misread "possible" as "positive"
The inverse function of xe^x. You plug in some value and it shows you what x was used.
Holy shit end whatever academic career you have and get the fuck off Veeky Forums
1.) x^(2) = 2^(x)
2.)ln[x^(2)] = ln[2^(x)]
3.) 2ln(x) = xln(2)
4.) ln(x) = x*ln(2) * 1/2
5.) ln(x) * 1/ x = ln(2) * 1/2
*Note* I used common sense here as a tool
6.) x = 2, x = 4. Only 2 answers, OP is a faggot.
>6.) x = 2, x = 4. Only 2 answers, OP is a faggot.
why does the graph give 3 answers, just wondering
Yes, if you are looking for fraction answers, sure. I'll admit that I'm wrong, as I was looking for whole numbers only.
Ln (x^2) = 2ln|x|...!
that third solution is what needs Lambert W, your solution is incomplete. Not too bad but still needs work on why are only these 2 and there are not more whole solutions, then ask yourself if there are other solutions that are not whole and leave it as an open problem because it surpasses your current tools.
What I'm trying to say is, you didn't use your tools (log) as exhaustively as you should, as a mathematician.
thanks for this post
OP equation looked interesting and this is the answer, lambert w function
en.wikipedia.org
learned something new today that I will probably never again use in my life