# How do i solve the following with or without use of logs?

1. Oct 3, 2009

### Dell

How do i solve the following with or without use of logs?
3^x=x^2

the problem being that i have x both in my exp and in my base, what i have done is taken each side and gotten

(3^x)^(1/x)=(x^2)^(1/x)
3=x^(2/x)
3^(1/2)=(x^(2/x))^(1/2)
sqrt(3)=x^(1/x)

0.5*ln(3)=(1/x)*ln(x)

2. Oct 3, 2009

### LCKurtz

You can't solve it algebraically. It can be solved in terms of a non-elementary function called the Lambert W function. Numerically the answer is -.6860267245.

3. Oct 3, 2009

### Dell

on my calculator i can get x=1.0434693552 but cannot prove it, is there no other way other than Lambert W function?

4. Oct 3, 2009

### LCKurtz

No. That's it. Numerically, which is how the Lambert W function's values are gotten anyway. And your x=1.0434693552 doesn't solve 3x = x2.

5. Oct 3, 2009

### Dell

no sorry my mistake, that was for another excercise,
i looked up lambert W functions and i got this

3x=x2
(31/2)x=x
1=x/(31/2)x =====> (31/2)=a
1=x/ax
1=x*e-x*ln(a)
-ln(a)=[(-ln(a)*x)*e(-ln(a)*x)

after applying W to each side

-ln(a)*x=W(-ln(a))
x=-W(-ln(a))/(ln(a))

is this correct, and also how do i get a numerical answer for this??

Last edited: Oct 3, 2009
6. Oct 3, 2009

### Dell

i found an online calculator that gave me the value, which looks right, but it only gives me one value and i believe there are 2. how can i work it out?

7. Oct 3, 2009

### Mentallic

No, there is only 1 real value.