# Why is this?

1. Dec 13, 2006

### kasse

Why is it that ln(x^2+1) = ln (abs: 2y-3) can be transformed into x^2+1 = 2y-3 , this according to an example in my book.

What I don't understand is why you can eliminate the absolute sign.

2. Dec 13, 2006

### marlon

Huh ? The argument of the ln function always needs to be positive by definition : if ln(A) = B then exp(B) = A.

marlon

3. Dec 13, 2006

### kasse

Last edited by a moderator: Apr 22, 2017
4. Dec 13, 2006

### marlon

That's Norwegian, right ?

yep, they write the absolute sign because the argument of the ln function NEEDS to be positive. So in that case, you only consider the y values for which 2y-3 > 0 --> y>1.5

marlon

5. Dec 13, 2006

### kasse

I don't really understand why it needs to be positive. Is it not possible to say that e^(ln(abs: 2y-3)) = (abs: 2y-3) ?

And yes, it's Norwegian. Did you guess that without checking my profile?

Last edited: Dec 13, 2006
6. Dec 13, 2006

### marlon

e^(ln(abs: 2y-3)) = (abs: 2y-3) : YES THAT IS CORRECT.

Now, if you are using the absolute value function like |x| and you know that x > 0 (because of the argument of ln), the definition of |.| says that |x| = x.

In short : definition for |.|

|x| = x if x > 0
|x| =-x if x < 0

So |2y-3| = 2y-3 IF 2y-3 > 0

Finally, let me repeat why the argument (2y-3) of the ln function is positive BY DEFINITION:

ln(A) = B <--> A = exp(B)

But exp(B) is always positive so A must always be positive.

Keep in mind that exp(B) = $$e^B = (2.7)^B$$

When you exponentiate 2.7, it can never become a negative number.

Yes, my native language is Dutch and some of the words look very much alike.

marlon

Last edited: Dec 13, 2006
7. Dec 13, 2006

### HallsofIvy

Staff Emeritus
The citation you give doesn't say that! It says "if ln(x2+ 1)= ln(|2y- 3|)+ C1, then x2+ 1= C1(2y-3)".

You don't need the "| |" because the C1 can be chosen positive or negative.