Solving Equation With Natural Log

[darshanpatel]

Homework Statement

Solve the Equation, Check for extraneous solutions.

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

Homework Equations

-None-

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

Please help! Thanks

 

[micromass]

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

 

[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)

?

 

[Mark44]

What do you mean quadratic equation?

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.

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)

?

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.

 

[darshanpatel]

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

 

[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)

 

[HallsofIvy]

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.

 

[Joffan]

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

 

[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.

 

[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?

 

[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?

 

[SammyS]

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?

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

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

 

[darshanpatel]

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

e^ln(x)=e^2
x=e^2
x is about 7.389

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

x= 1, 7.389

right?

 

[darshanpatel]

Would that be the correct answer?

 

[BloodyFrozen]

Yes, that is correct.

http://www.wolframalpha.com/input/?i=(ln x)^2 = ln (x^2)