# Solving Equation With Natural Log

1. Jan 20, 2012

### darshanpatel

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

Solve the Equation, Check for extraneous solutions.

(ln x)^2 = ln (x^2)

2. Relevant equations

-None-

3. The attempt at a solution

I tried simplifying it down for myself but got stuck and don't know what to do next.

(ln x)^2= ln x^2

(loge x)^2 = loge x^2

2. Jan 20, 2012

### micromass

Staff Emeritus
Use $ln(x^2)=2ln(x)$ to obtain a quadratic equation in ln(x).

3. Jan 20, 2012

### darshanpatel

What do you mean quadratic equation?

I know through what you are saying that I will get 0= ((ln x)^2)-2lnx right?

would i plug it into the quadratic forumla or something like:

x= (-(2ln) +/-√((2ln)^2 - 4(lnx)(0)))/2(lnx)

?

4. Jan 20, 2012

### Staff: Mentor

The equation is actually quadratic in ln(x). IOW, that means it's an equation in which ln(x) appears to, at most, the second power.
You wouldn't be solving for x; you will be solving for ln(x), and you should get two values. For each value you can solve the resulting equation for x.

Note that 2ln is meaningless, just as the symbol √ by itself is also meaningless.

5. Jan 20, 2012

### darshanpatel

How do I solve for it? What would it be/look like?

6. Jan 20, 2012

### daric soldar

You had the right idea going, darshanpatel. If you follow Mark44's advice and instead of using the quadratic equation to find x, try to find ln(x). In your case, if you replace ln(x) = y or some other variable into your first equation in post #3, you would get:

y^2 - 2y = 0

If you solve for y using the Quadratic Equation, you should get two answers.

After you get those two answers, replace back in ln(x) for y.

Then solve for x with your two equations.

Good luck!

(I hope that made sense and uses correct logic. I am an engineering major, so I am trained toward using mathematics as a tool for design purposes. I love using math; I just am not trained in proofs or at in-depth mathematics period....at least, not yet)

7. Jan 21, 2012

### HallsofIvy

Staff Emeritus
You surely don't need the quadratic formula to solve $y^2- 2y= 0$! However, although the quadratic equation has two solutions, the original equation has only one solution.

8. Jan 21, 2012

### Joffan

I get two perfectly viable solutions for the original equation from this process, HallsofIvy.

9. Jan 21, 2012

### daric soldar

You do get two solutions, HallsofIvy. You were correct, however, in saying that the Quadratic Formula is not necessary to solve the equation.

10. Jan 21, 2012

### darshanpatel

so would it be like this:

(ln x)^2=ln x^2

(ln x)^2= 2ln x

ln(x)=y

y^2=2y

y^2-2y=0

y(y-2)=o

y=0 and y=2

as solutions?

because if i plug this into the quadratic equation:

y^2-2y=0

where a:1 b:-2 and c: 0

I still get: y=0 and y=2

is that right?

11. Jan 21, 2012

### darshanpatel

After I get y=0 and y=2, i plug it into lnx^2-2lnx=0 right?

So plug in 0 and 2 for the x's?

12. Jan 21, 2012

### SammyS

Staff Emeritus
You don't plug 0 and 4 in for x !

You solve each equation for x: ln(x) = 2 and ln(x) = 0 .

13. Jan 22, 2012

### darshanpatel

so would it be like for ln(x)=2

e^ln(x)=e^2
x=e^2

and ln(x)=0
e^ln(x)=e^0
x=1

x= 1, 7.389

right?

14. Jan 22, 2012

### darshanpatel

Would that be the correct answer?

15. Jan 22, 2012