# Thread: Constant plus natural log of constant. Find constant?

1. ## Constant plus natural log of constant. Find constant?

On the title, I meant variable.

I need to know how to solve for a variable when you have a variable plus a number multiplied by the natural log of the same variable, and it equals some incredibly complex number. It's constant, as I stated, so that side is pointless and can be treated as c.

so... how do I solve for x in a situation such as...

Make somenumber anything you want. Mine specifically is is between 8 and 9. It is just the quantity of a number plus a number multiplied by the natural log of another number end quantity all over 6. Ignoring that, could anyone help?

This is actually something simple I think I should know. I am just trying to wrap up a much more complicated calculus problem. I just forgot how to do this and cannot seem to find it anywhere.

2. $x^{2}+2ln(x)=a$ is kind of difficult to solve for x algebraically.

You can always try Newton's method to solve for a particular value of x.

Suppose you had $x^{2}+2ln(x)=3$

Try an initial guess of 1.5

$1.5-\frac{(1.5)^{2}+2ln(1.5)-3}{2(1.5)+\frac{2}{1.5}}=1.485939...$

Keep iterating unitl you hone in. This converges pretty fast, so not many are needed to get it within a few decimal places.

3. Have you ever heard of the LambertW function?. That can be helpful when it's difficult to solve algebraically.

For instance, to solve your $y=x^{2}+2ln(x)$ for x, we get:

$x=e^{\frac{-1}{2}LambertW(e^{y})+\frac{y}{2}}$

4. Thanks. I am going to go with the first way you explained it, though.

5. I have not had a lot of experience with it either. But, I can show you a basic example. I will make one up.

Something like $4^{t}=9t$. Solve for t.

The idea is to get it into the form $A=Be^{B}$

$4^{t}=9t$

$1=\frac{9t}{4^{t}}$

$1=9te^{-tln(4)}$

$\frac{1}{9}=e^{-tln(4)}$

$\frac{-ln(4)}{9}=(-tln(4))e^{-tln(4)}$....required form A=Be^B

$-tln(4)=W(\frac{-ln(4)}{9})$

$t=\frac{W(\frac{-ln(4)}{9})}{ln(4)}$

As far as getting a numerical value for this, I think there is a table you look them up in.

This case works out to be t=2.130588... or t=.1337459....

6. What I have is

$
c^2+3ln(c)=\frac{20+27ln(3)}{6}
$

I am on the average value of a function, which is quite easy... The right side is the average value with certain bounds.
The left side is the function as translated into c, where c = a place where the functions hits it. That function being:

$
f(x)=x^2+3ln(x)
$