I have been trying very hard to solve this equation without a calculator, but to no avail. I learnt about Lambert W function and using it I reached x < W(5x + xlnx). when I asked my math teacher about he said he has no idea, I hope someone can help.
#An equation I have been trying to solve for a long time
tepid moat
alpine saffronBOT
I have been trying very hard to solve this equation without a calculator, but to...
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still cloud
I have been trying very hard to solve this equation without a calculator, but to...
what are u trying to find?
all x, or..?
tepid moat
the answer is 0.018664 < x < 1.71152, but I actually want to know how the work has been done
and no, I don't want graphing, nor newton's method
I want actually an exact answer
still cloud
the answer is 0.018664 < x < 1.71152, but I actually want to know how the work h...
it hasn't
WA numerically approximated it
tepid moat
WA numerically approximated it
yeah, I mean an answer with Lambert W is accepted
still cloud
define $f$ such that $f^{-1}(x)=e^x-\ln x$
lapis arrowBOT
fr
tepid moat
Sorry man, but I dont get what you mean
but what I mean is, I want an answer that would produce an exact number
tepid moat
it can have logarithms and lambert W np
still cloud
it can have logarithms and lambert W np
yeah then just define ur very own function
and say that's the answer
if ur acceptig lambert W like that
tepid moat
yeah then just define ur very own function
where R(xe^e^x) = x?
i mean smt like that?
rapid cedar
I have been trying very hard to solve this equation without a calculator, but to...
...why'd you block part of the problem?
tepid moat
...why'd you block part of the problem?
This is actually work from past papers
we do work on classkick
and this is a screenshot
the teacher removed the classkick and what I have here is only the screenshot
as for why, I am not sure why I did
tepid moat
uhhh
I can show you the two statements I reached to
but as for how, I could if you actually need that
rapid cedar
Okay, I see, the problem is that you didn't try to isolate x.
And I'm not sure if you can.
I don't think you can actually generate the term needed for a Lambert W function.
Because you have e^x and ln(x).
tepid moat
Okay, I see, the problem is that you didn't try to isolate x.
I did in scattered attempts in the last two months
rapid cedar
I did in scattered attempts in the last two months
No, I mean in the thing that you posted here. You just multiplied both sides by x.
tepid moat
No, I mean in the thing that you posted here. You just multiplied both sides by ...
yes, that is try to isolate the X
I also reached to e^e^x < e^5 * x
though still of no use
rapid cedar
yes, that is try to isolate the X
...you isolate the x by introducing it to both sides when it's already on both sides?
tepid moat
...you isolate the x by introducing it to both sides when it's already on both s...
That actually can help in some situations
I mean like, to solve 1/x < x you need to multiply both sides by x
but that is beside the point
rapid cedar
I mean like, to solve 1/x < x you need to multiply both sides by x
...yes, because that removes x from one side.
tepid moat
whatever it is, I tried my best, and that is what I got
can you provide us with help, if yes, then do
rapid cedar
Hmm. Well, e^(-x) (5 + ln(x)) > 1.
tepid moat
Hmm. Well, `e^(-x) (5 + ln(x)) > 1`.
yeah that is what you do clasically with LambertW function problems
rapid cedar
Oh hey. e^(-x) (5 + ln(x)) - 1 > 0.
Which means that we're looking for roots of the function f(x) = e^(-x) (5 + ln(x)) - 1.
tepid moat
alright
rapid cedar
So maybe we can at least narrow down our search space by differentiating?
I mean, if it has extrema on either side of the x-axis then we know a root exists between them by the intermediate value theorem.
tepid moat
So maybe we can at least narrow down our search space by differentiating?
Sorry if my way of explaining what I want wasn't clear, at the end I am new to these stuff.
I know the answer when it comes to numbers, I know the numerical answer
I want to solve it analytically, I want an answer that is exact, like maybe W(5ln(e^e)) or smt
rapid cedar
,w e^x > 5 + ln(x)
lapis arrowBOT
tepid moat
I actually scoured the internet to find an answer
lapis arrowBOT
tepid moat
Maybe for an equation?
same
rapid cedar
I actually scoured the internet to find an answer
...I really think you mean "scoured".
tepid moat
actually "scattered" what I meant lol
rapid cedar
actually "scattered" what I meant lol
I don't think that's correct either.
rapid cedar
Fair.
tepid moat
anyways
I think maybe what can help solve the problem is trying to find solutions to e^e^x = a + x
this guy did actually solve something that could really help us, though sadly he assumed that e^(x-a) and lnx touch on one point, thus that they are tanjet when they meet
rapid cedar
this guy did actually solve something that could really help us, though sadly he...
*tangent. And yeah, that was his goal. He wanted them to be tangent.
Okay, so I don't know how this works for the inequality, but I have a path forward if we're solving an equation, I think.
rapid cedar
Okay, so. e^x = 5 + ln(x), right?
tepid moat
yes
rapid cedar
Therefore x = ln(5 + ln(x)).
tepid moat
yes
rapid cedar
...oh wait, no, that's not gonna help at all. That's just gonna make everything worse.
tepid moat
wait
rapid cedar
Because obviously, if we plug that value in on the left, we just get a trivial equation. My idea was to plug it in on the right, but that creates a greater divide between the x's.
tepid moat
hmmm
rapid cedar
Wait!
tepid moat
wait
calculus?
because we can plug the x inside
and again
and again
and see the pattern
that might help
rapid cedar
I mean, sure. We can get an infinite family of equations.
I don't know if we can get one that's more solvable, though.
tepid moat
I mean, sure. We can get an infinite family of equations.
man
ok I tried it on desmos
pluged it into itself mutiple times
one of the answers is showing up
I am sure calculus can help here somehow
the two answers given by the wolf math website ( I dont remember its name ) are x ≈ 0.0186639620506490... and x ≈ 1.71152201393235...
rapid cedar
the two answers given by the wolf math website ( I dont remember its name ) are ...
...Wolfram Alpha?
tepid moat
...Wolfram Alpha?
Yes
hexed frigateBOT
@tepid moat
tepid moat
+close
hexed frigateBOT
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