SRL Archive - Mathematics

e^x*lnx = 1

I've been trying to solve that but haven't succeeded.
Anyone wanna give it a shot? :p

So this is the same as

ln(x)*e^x = 1 right?
That means ln(x)*e^x - 1 = 0

Put that in my TI-84, and the answer comes out to be x = 1.310

I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.

From: 3Garrett3
So this is the same as

ln(x)*e^x = 1 right?
That means ln(x)*e^x - 1 = 0

Put that in my TI-84, and the answer comes out to be x = 1.310

I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.

Hehe, yeah I know how to approximate roots but I'd like to know the exact answer.

From: 3Garrett3
So this is the same as

ln(x)*e^x = 1 right?
That means ln(x)*e^x - 1 = 0

Put that in my TI-84, and the answer comes out to be x = 1.310

I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.

why do you have to subtract the one to use it in the calculator? you shouldnt have to do that :p

and isnt ln(x) the reverse of e^x?

From: Awkwardsaw
why do you have to subtract the one to use it in the calculator? you shouldnt have to do that :p

and isnt ln(x) the reverse of e^x?

Well, my calculator is the best that IB allows, and they don't allow something where you can just enter the equation into it and it tells you the exact answers and how to get it. I have a *Solver* function, which must be equal to 0 in order for it to be used, and it only gives approximate answers.

And ln(x) is the log(base e)x. I know that ln(e^x)) is simply x. But when they are multiplied together, I don't think the same rules apply.

From: 3Garrett3
Well, my calculator is the best that IB allows, and they don't allow something where you can just enter the equation into it and it tells you the exact answers and how to get it. I have a *Solver* function, which must be equal to 0 in order for it to be used, and it only gives approximate answers.

And ln(x) is the log(base e)x. I know that ln(e^x)) is simply x. But when they are multiplied together, I don't think the same rules apply.

*cough*get a casio*cough*

and kay, i havnt used logs for ever lol. but it seems like there should be somesort of cancelation(sp)?

From: Awkwardsaw
*cough*get a casio*cough*

and kay, i havnt used logs for ever lol. but it seems like there should be somesort of cancelation(sp)?

I was provided with a free TI-84 Plus just for doing IB. I wouldn't spend money on a calculator which either a) wasn't allowed by IB or b)could be upgraded when I got to uni.

I'm not sure yet what kind of calculator I'll be allowed in my engineering courses next year, so I wont be getting one until then.

But ontopic, my least favourite portion of the math curriculum was the logs and e^x, so I may have forgotten a rule about cancellations, but I don't remember there being one when they were multiplied.

From: 3Garrett3
I was provided with a free TI-84 Plus just for doing IB. I wouldn't spend money on a calculator which either a) wasn't allowed by IB or b)could be upgraded when I got to uni.

I'm not sure yet what kind of calculator I'll be allowed in my engineering courses next year, so I wont be getting one until then.

But ontopic, my least favourite portion of the math curriculum was the logs and e^x, so I may have forgotten a rule about cancellations, but I don't remember there being one when they were multiplied.

e^lnx = x

But if you like Texas Instruments like I do, get a TI-89, it's pure awesomeness! :)
http://education.ti.com/educationportal/sites/US/productDetail/us_ti89ti.html

From: marpis
e^lnx = x

But if you like Texas Instruments like I do, get a TI-89, it's pure awesomeness! :)
http://education.ti.com/educationportal/sites/US/productDetail/us_ti89ti.html

I do like them, but I am not sure what my Uni is going to say, until I get there next year. So I'll be safe rather than sorry.

e^lnx = x as well as ln(e^x) = x correct?

Indeed interesting equation.. I've failed to solve it :(.. Curious to see the answer!

From: 3Garrett3
I do like them, but I am not sure what my Uni is going to say, until I get there next year. So I'll be safe rather than sorry.

e^lnx = x as well as ln(e^x) = x correct?

yep.

is x ln x power or its (e^x)(lnX)


lol here is my poor attempt

line 8??? log x = 10^x ????

yeah e^x cancels out.

Besides my answer was obviously wrong, but a attempt wasn't bad :).

I mean… at line 8 you rewrite log x as 10^x…. which isn't even true (e.g. x =1 ==> 0 = 10)

Yeah, I did them while half a sleep lol.
I showed my teacher this, she couldn't do it. I will attempt it again.

Zomg… just use netwon's method :)

From: Main_Ftw

Zomg… just use netwon's method :)

If you want an approximation zomg just use texas instrument :p

Exact answer:
http://mathbin.net/46986

Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!

:)

From: Nava2
Exact answer:
http://mathbin.net/46986

Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!

:)

Oh, but I see 2 variables there. That's not an exact answer, nor an answer at all, just a different way to express the function.
And, before anyone dares to say it has no exact answer, here's a proof there is one:
ln(x) is continous and deriving ]0, infinity[
e^x is continuous and deriving ]-inf, inf[
-> ln(x)*e^x is continous and deriving ]0, infinity[
thus lnx*e^x = 1 has a solution (only one in this case, derivative is always positive)

From: marpis
Oh, but I see 2 variables there. That's not an exact answer, nor an answer at all, just a different way to express the function.

Actually it is an answer ;)! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.

In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.

From: Nava2
Actually it is an answer ;)! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.

In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.

You see, your sum has both "n" and "x" there, but only giving values to "n".
How can you say it gives a number as an answer o_O Where does the x go?

And I don't want an approximation :p I want an answer like
"e^x*lnx = 1 when x = e^2*ln(2)" or something like that.

From: marpis
You see, your sum has both "n" and "x" there, but only giving values to "n".
How can you say it gives a number as an answer o_O Where does the x go?

And I don't want an approximation :p I want an answer like
"e^x*lnx = 1 when x = e^2*ln(2)" or something like that.

If you would read what I said, you would see there is an exact answer as a sum of fractions.

From: Nava2
Exact answer:
http://mathbin.net/46986
Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!
:)

^ There's just the infinite sum, which is not an answer, but another way to express the function.

From: Nava2
Actually it is an answer ;)! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.

In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.

^ We used the answer to calculate the answer? You only posted a way to find an approximation

From: Nava2
If you would read what I said, you would see there is an exact answer as a sum of fractions.

^ Sorry but I don't understand.

http://en.wikipedia.org/wiki/Alternating_series

From: Nava2
http://en.wikipedia.org/wiki/Alternating_series

:rolleyes:

You see the difference between the series you presented as a solution to the function e^x*lnx and these?

The first one has 2 variables: n, which is given values, but then there's x which gets no values.
In the wikipedia article you posted there is only one variable in the function, n or k in some.

It's a function if it has variables that are not given any values.
It's a number if all variables get known values.

From: Nava2
http://en.wikipedia.org/wiki/Alternating_series

:rolleyes:

You see the difference between the series you presented as a solution to the function e^x*lnx and these?

The first one has 2 variables: n, which is given values, but then there's x which gets no values.
In the wikipedia article you posted there is only one variable in the function, n or k in some.

It's a function if it has variables that are not given any values.
It's a number if all variables get known values.
I want a numeric answer to an equation.

Next question!

e^xlnx = 1
lnx = 1/(e^x)

1/(e^x) = ln1 - ln(e^x)
= - ln(e^x)
= - x lne
= -x
therefore

lnx = -x
lnx + x = 0
x + e^x =0

that's as far as I got it, I think I messed something up in my substitution in the second line

anyone found the answer?

Sorry if it looks bad this iam on my phone I will edit later

From: Bonfield
e^xlnx = 1

Need to be careful, the equation in the first post isn't clear.

The actual equation, which you can see by following the link is :- ln(x)*e^x = 1.

Edit :- Made a mistake, thanks for pointing that out Method!

From: Bonfield

e^xlnx = 1
lnx = 1/(e^x)
1/(e^x) = ln1 - ln(e^x)
= - ln(e^x)
= - x lne
= -x
therefore

lnx = -x
lnx + x = 0
x + e^x =0

that's as far as I got it, I think I messed something up in my substitution in the second line

anyone found the answer?

Sorry if it looks bad this iam on my phone I will edit later

On the bolded part you're saying that ln1 - ln(e^x) = lnx
even though ln1=0 and ln(e^x)=x -> ln1 - ln(e^x) = 0 - x = -x
lnx <> -x
:(

From: Aequix
Need to be careful, the equation in the first post isn't clear.

That equation would have a simple solution (As far as I can tell) :-

e^(xln(x)) = 1

As e^ln(x) = x, it follows that e^(xln(x)) = x^2.

Therefore, x^2 = 1 and as such the solution is x = +/- Sqrt(1)

The actual equation, which you can see by following the link is :- ln(x)*e^x = 1.

And the first equation is very simple if you understand calculating order.
e^x3/4 = (e^x)(3/4)
e^(x3/4) = e^((3x)/4) = e^((3/4)x)
lnx3+1 = 3ln(x)+1
ln(x+3)+1 = ln(x+3)+1
lnxe = e*ln(x)
lnxe^x = ln(x)e^x

You can't just add brackets however you want to, leaving them off is just as important as adding them.

Naturally, simply stating that there's been a few slip-ups throughout the thread due to the way that the equation is read. Following rules you'll come out fine, but a lot of people tend to find bracketing can clarify most-equations better than leaving them out.

From: Marpis

You can't just add brackets however you want to, leaving them off is just as important as adding them

Given the situation of-course. ;)

From: Aequix
Need to be careful, the equation in the first post isn't clear.

That equation would have a simple solution (As far as I can tell) :-

e^(xln(x)) = 1

As e^ln(x) = x, it follows that e^(xln(x)) = x^2.

Therefore, x^2 = 1 and as such the solution is x = +/- Sqrt(1)

The actual equation, which you can see by following the link is :- ln(x)*e^x = 1.

e^(xln(x)) is x^x, not x^2, which makes the problem more complex.

On the bolded part you're saying that ln1 - ln(e^x) = lnx
even though ln1=0 and ln(e^x)=x -> ln1 - ln(e^x) = 0 - x = -x
lnx <> -x
:( [\QUOTE]

why can't (e^x)^-1 equal -x
here

Should I take the log of the other side ie ln(lnx) = -x ? I don't think that is right either :( I don't know if that's possible to solve, its been so long since I have done any of this stuff, I will sit down and have a proper go at this tomorrow

just curious do you have the answer marpis? And where did you getthis question?

From: Bonfield

On the bolded part you're saying that ln1 - ln(e^x) = lnx
even though ln1=0 and ln(e^x)=x -> ln1 - ln(e^x) = 0 - x = -x
lnx <> -x
:( [\QUOTE]

why can't (e^x)^-1 equal -x
here

Should I take the log of the other side ie ln(lnx) = -x ? I don't think that is right either :( I don't know if that's possible to solve, its been so long since I have done any of this stuff, I will sit down and have a proper go at this tomorrow

just curious do you have the answer marpis? And where did you getthis question?

No, I dont and I just came up with it :D Didn't hear about it anywhere, I just like inspecting random functions with "e" in them when I'm bored. Go ahead, call me a nerd.

From: marpis


No, I dont and I just came up with it :D Didn't hear about it anywhere, I just like inspecting random functions with "e" in them when I'm bored. Go ahead, call me a nerd.

Nah a nerd would except the answer Nava2 gave you ;)

ok, this is a very interesting equation.

if you are asking what is the work required to find the answer then…the answer is that there is no answer. the computer doesnt calculate this algebraicly. it uses a nice guess and check method. your equation is also equal to ln(ln(x))=-x :]

that is the most simplified that equation can get :\

How is ln(ln(x)) = -x simplified compared to the original equation?

ln(lnx) + x = 0 is not making the equation easier it pretty much just been re-written and I tried to solve it that's as far as I could try to simplify it

I tried it with x = 1.31 as Garrett posted earlier and it does work, I'm just unsure as to what the next step is

the equation is not simpler or harder. it is the same. you are right. that is as simplified as it gets. this cant be solved algebraicly

in other words there is no next step. there are a ton of equations which cannot be solved algebraicly. just like there is no formula for PI

Ok u just thought yo could take it further, so now the only way to find the answer is an approximation

lol. that is as far as it goes. it might be an irrational number. ii am fairly sure it is. but you could guess and check to see if you can reach the end. i would run a program that did it. never do it myself :P. i think it is likely that it might never end though.

From: g0tp0t
the equation is not simpler or harder. it is the same. you are right. that is as simplified as it gets. this cant be solved algebraicly

in other words there is no next step. there are a ton of equations which cannot be solved algebraicly. just like there is no formula for PI

Except that there are many many many formulas for pi, mostly endless sums.
Examples:

From: g0tp0t
lol. that is as far as it goes. it might be an irrational number. ii am fairly sure it is. but you could guess and check to see if you can reach the end. i would run a program that did it. never do it myself :P. i think it is likely that it might never end though.

You can't "reach the end" because computers only handle a limited amount of decimals, and when that is reached it says it has found a root eventhough it's really not.
The answer will most likely have e in it, which makes it irrational and transcedent, but that's not a problem ofcourse.

lol.yea. see how long it takes to get this to 1k digits :P

My teacher said before that pi can be derived from limit as n -> infinity of polygon's perimeter divided by 2 or something. n is the number of sides.

From: Main_Ftw
My teacher said before that pi can be derived from limit as n -> infinity of polygon's perimeter divided by 2 or something. n is the number of sides.

Yea that's a very efficient algorithm, we went through that in maths.

ln(x) is continous and deriving ]0, infinity[
e^x is continuous and deriving ]-inf, inf[
-> ln(x)*e^x is continous and deriving ]0, infinity[
thus lnx*e^x = 1 has a solution (only one in this case, derivative is always positive)

Are you sure?

3 is continous and deriving ]0, infinity[
5 is continuous and deriving ]-inf, inf[
-> 3*5 is continous and deriving ]0, infinity[
thus 3*5 = 1 has a solution (only one in this case, derivative is always positive)

Clearly, 3*5 does not equal 1…

From: lollage22
Are you sure?

3 is continous and deriving ]0, infinity[
5 is continuous and deriving ]-inf, inf[
-> 3*5 is continous and deriving ]0, infinity[
thus 3*5 = 1 has a solution (only one in this case, derivative is always positive)

Clearly, 3*5 does not equal 1…

They are inverse functions though, so you would have to compare inverses.

8*(1/8)=1

No, i think it is


you need those limits, and continuity. A proof why the solution is not a fraction shouldn't be too hard. Furthermore, the answer could be defined as the solution of the equation, and so be definable as an infinite sum. The matter of being eachother inverses is not needed here.