Solve e^x=ln(x)

  • #1

Homework Statement



solve e^x=ln(x)


Homework Equations



e^x=ln(x)


The Attempt at a Solution



I'm not sure where to start. I've been thinking of Euler's but can't quite get to it.
I'm sure there is an easier solution.

The answer will surely be a complex number.
 

Answers and Replies

  • #2
6
0
e^x is inverse of ln(x).Both these functions are symmetric about y=x..you can draw graphs
of both to see that...There is no real point of intersection in these graphs...thus according to me there must not be any solution..!
 
  • #3
Are you talking about real numbers?
I think the solution will be a complex number. You know, y=x+iz

I can't plot that by hand. Do you have a tool that takes complex numbers?
 
  • #4
6
0
i know complex numbers..but one cannot plot such curves on argand's plane...you cannot plot even (x,y), where both are real quantities..isnt it..?? so how do you propose to plot these entire curves..
 
  • #5
1,800
53
You can draw them. Now, how about this: If you don't have Mathematica, try and find a machine running it and then try and interpret my code or I mean you really can just look at it and see what I'm doing right?

Code:
ContourPlot[{Exp[r*Cos[t]]*Cos[r*Sin[t]] - 
     Log[r] == 0, Exp[r*Cos[t]]*Sin[r*Sin[t]] - 
     t == 0}, {r, 0, 10}, {t, 0, 2}]

Look at plot and see where they cross to get a starting point for the numerical root finder then:


Code:
In[140]:=
{ra, ta} = {r, t} /. FindRoot[
     {Exp[r*Cos[t]]*Cos[r*Sin[t]] - Log[r] == 0, 
      Exp[r*Cos[t]]*Sin[r*Sin[t]] - t == 0}, 
     {{r, 1.341}, {t, 1.319}}]; 

Exp[r*Exp[I*t]] /. {r -> ra, t -> ta}
Log[r*Exp[I*t]] /. {r -> ra, t -> ta}

Out[141]=
0.31813150520476396 + 1.3372357014306893*I

Out[142]=
0.31813150520476413 + 1.3372357014306895*I
 
Last edited:
  • #6
@ jackmell
Thank you!

I used www.wolframalpha.com and got:
0.3181... + 1.337235...*I
0.3181... - 1.337235...*I
Please note the (-) on the second root. I'll try to figure out why that is.

Could you please show the step to get to the ekvation system:
Exp[r*Cos[t]]*Cos[r*Sin[t]] - Log[r] = 0,
Exp[r*Cos[t]]*Sin[r*Sin[t]] - t = 0

If you don't have the time, then; thank you for your help so far.


@ xxhizors
It was you who suggested plotting them...
 
  • #7
1,800
53
@ jackmell
Thank you!

I used www.wolframalpha.com and got:
0.3181... + 1.337235...*I
0.3181... - 1.337235...*I
Please note the (-) on the second root. I'll try to figure out why that is.

Could you please show the step to get to the ekvation system:
Exp[r*Cos[t]]*Cos[r*Sin[t]] - Log[r] = 0,
Exp[r*Cos[t]]*Sin[r*Sin[t]] - t = 0

...

That's just the real and imaginary parts of e^z and log(z) set equal to each other right? Let z=re^{it} and then expand e^z and log(z) into their real and imaginary parts. So:

[tex]\log(z)=log(r)+it[/tex]

[tex]e^z=\exp\bigg(re^{it}\bigg)=\exp\bigg(r(\cos(t)+i\sin(t)\bigg)[/tex]

finish expanding that out. And keep in mind log is log to base e in complex analysis (and in Mathematica) and not that 10 thing like in high-school.
 
  • #8
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
@ jackmell
Thank you!

I used www.wolframalpha.com and got:
0.3181... + 1.337235...*I
0.3181... - 1.337235...*I
Please note the (-) on the second root. I'll try to figure out why that is.

Could you please show the step to get to the ekvation system:
Exp[r*Cos[t]]*Cos[r*Sin[t]] - Log[r] = 0,
Exp[r*Cos[t]]*Sin[r*Sin[t]] - t = 0

If you don't have the time, then; thank you for your help so far.


@ xxhizors
It was you who suggested plotting them...

The Maple 14 commands sol:=solve(exp(z)=log(z),z): s:=allvalues(solz) give 7 solutions:
.31813150520476413531-1.3372357014306894089*I
1.3307856523213187947-31.831890458971534280*I
1.2758333591184965919-25.571310933744525843*I
1.3739868937005621738-38.098047998110211146*I
1.0899596304118028315-13.090220124707805661*I
1.2014904900359295887-19.320504273195991825*I
.88399815002006774174-6.9222760868159367889*I
where I = sqrt(-1). Of course, the 7 complex conjugates of the above are also roots, so we get 14 roots altogether.

RGV
 
  • #9
1,800
53
It's more than that right? I think there are an infinite number of solutions but don't know how to prove that. Here's some along the real axis (where the two curvers cross) and they repeat by 2pi along the imaginary axis I believe.

Anyone interested in proving how many there are?
 

Attachments

  • theroots.jpg
    theroots.jpg
    25.1 KB · Views: 1,125

Related Threads on Solve e^x=ln(x)

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
14
Views
7K
  • Last Post
Replies
9
Views
39K
  • Last Post
Replies
9
Views
17K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
15
Views
5K
  • Last Post
Replies
10
Views
46K
Top