How to Solve for x in ln(x) + x = 10: Quick and Easy Method

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SUMMARY

The discussion focuses on solving the equation ln(x) + x = 10 using numerical methods. Participants confirm that an exact solution is not feasible, but Newton's Method is effective for approximation. The function f(x) = ln(x) + x - 10 is utilized, along with its derivative f'(x) = 1/x + 1. An initial guess of x0 = 1 leads to successive approximations, ultimately converging around x ≈ 7.929420095. The Lambert W function is also mentioned as a theoretical approach to express the solution.

PREREQUISITES
  • Understanding of logarithmic functions, specifically natural logarithms (ln).
  • Familiarity with Newton's Method for numerical approximation.
  • Basic knowledge of derivatives and their application in finding roots of functions.
  • Concept of the Lambert W function and its significance in solving transcendental equations.
NEXT STEPS
  • Study the implementation of Newton's Method in various programming languages.
  • Learn about the Lambert W function and its applications in solving equations.
  • Explore numerical methods for root-finding beyond Newton's Method, such as the Bisection Method or Secant Method.
  • Investigate the behavior of the function f(x) = ln(x) + x - 10 graphically to understand its roots.
USEFUL FOR

Mathematicians, students studying calculus, software developers implementing numerical methods, and anyone interested in solving transcendental equations.

j3n
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hello,
i heard there is some way of solving for x, but i can't for the life of me remember how. please help!

ln(x) + x = 10

thanks!
 
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Yeah, I think it should be something like this:
lnx>=1
x>=e...(1)
lnx>=1
lnx+x>=1+x
1+x<=10
x<=9
solution set for x: [e,9]
 
Graph intersection: y=ln x and y=10-x.
 
>.< I don't think that's it mate.

there's no way to solve for x, you can only approximate it.
 
Newton's method works quite nicely here

we will need, the function:

f (x) = ln(x) + x -10 = 0

and its first derivative:

f'(x) = 1/x + 1

an initial guess:

x0 = 1

and we can begin:

x1 = x0 - f(x0)/f'(x0) = 5.5

x2 = x1 - f(x1)/f'(x1) = 7.865213153

x3 = 7.929390693

x4 = 7.929420095

x5 = x4 as far as my calculator is concerned

try it out
 
Just curious how you would use Newton's Method, Gib Z, since you do not use a calculator.
Cheers,
 
Last edited:
j3n said:
hello,
i heard there is some way of solving for x, but i can't for the life of me remember how. please help!

ln(x) + x = 10

thanks!

I wonder if you are not thinking of the "Lambert W function". W(x) is defined as the inverse of the function f(x)= xex.

If ln(x)+ x= 10, then, taking the exponential of each side, eln(x)+ x= xex= e10.

x= W(e10).

Of course, the only way to evaluate that is to do some kind of numerical approximation as others have said,
 

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