- #1

- 3

- 0

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!!!

- Thread starter j3n
- Start date

- #1

- 3

- 0

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!!!

- #2

- 1,037

- 3

lnx>=1

x>=e...(1)

lnx>=1

lnx+x>=1+x

1+x<=10

x<=9

solution set for x: [e,9]

- #3

- 13,023

- 576

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

- #4

Gib Z

Homework Helper

- 3,346

- 5

>.< I don't think thats it mate.

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

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

- #5

- 63

- 0

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

- #6

- 184

- 0

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

Cheers,

Cheers,

Last edited:

- #7

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 961

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

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!!!

If ln(x)+ x= 10, then, taking the exponential of each side, e

x= W(e

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

- Last Post

- Replies
- 15

- Views
- 5K

- Last Post

- Replies
- 10

- Views
- 45K

- Replies
- 7

- Views
- 23K

- Last Post

- Replies
- 7

- Views
- 9K

- Last Post

- Replies
- 1

- Views
- 3K

- Last Post

- Replies
- 3

- Views
- 738

- Last Post

- Replies
- 4

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 1K