Understanding the Transformation of ln(x^2+1) = ln(abs: 2y-3) into x^2+1 = 2y-3

[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.

 

[marlon]

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

marlon

 

[kasse]

http://www.math.ntnu.no/emner/kode/SIF5003/gamle-eks/TMA4100_2006_08_18_lf.pdf

It's task 3 here. You won't understand the language, but hopefully understand the maths

 

[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

 

[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?

 

[marlon]

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) ?

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.

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

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

marlon

 

[HallsofIvy]

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.

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.