# Solving Natural Logarithms

1. Nov 1, 2011

### statusquorulz

1. The problem statement, all variables and given/known data

Solve for x

xln(x)-24+6ln(x)-4x=0

My attempt at the solution

I first moved "-24" and "-4x" to the right side of the equation yielding

xln(x)+6ln(x)=24+4x

I then converted the natural logarithms to exponent form and product form yielding

ln(x)^x+ln(x)^6 = 24+4x ====> ln(x^x*x^6)=24+4x =====> ln(x^(x+6))=24+4x

I then converted the equation in notation " e " form yielding

e^(24+4x)= x^(x+6)

Now I am stuck on where to proceed. I am aware that the answer has to be a positive number because the domain of a logarithm is always greater then 0.

Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 1, 2011

### HallsofIvy

Staff Emeritus
That is not going to have a solution in terms of any simple algebraic function. It might have a solution in terms of the "W" function which is defined as the inverse function to $f(x)= xe^x$.

3. Nov 1, 2011

### SteamKing

Staff Emeritus
You were on the right track when you collected like terms:
xln(x)+6ln(x)=24+4x

If you factor the LHS further, notice:
ln(x) (x+6)=24+4x

Can you figure out what x must be now? (Sorry, Mr. Lambert. No soup for you today.)