Solving e^x=ln(x): Tips and Tricks for Complex Number Solutions

  • Thread starter QuiteUnusual
  • Start date
In summary, the conversation discusses solving the equation e^x=ln(x) and the use of different methods such as Euler's method and plotting to find a solution. The solution is a complex number and there are multiple solutions, with an infinite number along the real and imaginary axes.
  • #1
QuiteUnusual
3
0

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.
 
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  • #2
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
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
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
QuiteUnusual said:
@ 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
QuiteUnusual said:
@ 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
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?
 

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What is e^x?

e^x, also known as the exponential function, is a mathematical function that represents the constant value 2.71828... raised to the power of x.

What is ln(x)?

ln(x), also known as the natural logarithm, is the inverse function of the exponential function. It represents the power to which e must be raised to equal x.

How do I solve e^x=ln(x)?

To solve this equation, you can use the Lambert W function, which is the inverse function of xe^x. Alternatively, you can use numerical methods such as graphing or iteration.

What is the solution to e^x=ln(x)?

The exact solution to e^x=ln(x) is x=W(1), where W is the Lambert W function. This solution cannot be expressed in terms of elementary functions, so it is often approximated using numerical methods.

What are the applications of e^x=ln(x)?

This equation has many applications in mathematics, physics, and engineering. For example, it can be used to model exponential growth and decay, as well as to solve differential equations. It also has applications in probability and statistics.

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