Ln(x) + x = c

[j3n]

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!

 

[chaoseverlasting]

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]

 

[dextercioby]

Graph intersection: y=ln x and y=10-x.

 

[Gib Z]

>.< I don't think that's it mate.

Theres no way to solve for x, you can only approximate it.

 

[mybsaccownt]

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

 

[theperthvan]

Just curious how you would use Newton's Method, Gib Z, since you do not use a calculator.
Cheers,

 

[HallsofIvy]

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)= xe^x.

If ln(x)+ x= 10, then, taking the exponential of each side, e^{ln(x)+ x}= xe^x= e¹⁰.

x= W(e¹⁰).

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