Why are there two solutions for e in ln| (y-2)/(y+2) | = 4x + c_2?

  • Thread starter Rijad Hadzic
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In summary, properties are important for understanding the natural world and are supported by a combination of evidence such as experimental data, observations, and mathematical proofs. In science, nothing is ever considered to be absolutely proven, but properties can be challenged or revised as new evidence is discovered. Understanding the underlying mechanism and how a property relates to others can provide deeper insight into the natural world and aid in making predictions. Overall, properties are essential for gaining a deeper understanding of the complex systems and processes that exist in our universe.
  • #1
Rijad Hadzic
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Homework Statement


Going from

[itex] ln|\frac {y-2} {y+2}| = 4x + c_2[/itex]

to

[itex] \frac {y-2} {y+2} = \pm e^{4x+c_2} [/itex]

Meaning, why is is [itex] \pm [/itex] e^{4x+c_2}?

Homework Equations

The Attempt at a Solution


I know that for log, you take the absolute value because ln(x) has restriction x>0

but I'm failing to see why its +- for e?
 
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  • #2
I think this has to do with the fact that the exponent is not restricted to the positive range; you can view this as the function within the log() only applies when it is positive, but without the log(), it can take up both positive and negative values.
 
  • #3
Rijad Hadzic said:

Homework Statement


Going from

[itex] ln|\frac {y-2} {y+2}| = 4x + c_2[/itex]

to

[itex] \frac {y-2} {y+2} = \pm e^{4x+c_2} [/itex]

Meaning, why is is [itex] \pm [/itex] e^{4x+c_2}?

Homework Equations

The Attempt at a Solution


I know that for log, you take the absolute value because ln(x) has restriction x>0

but I'm failing to see why its +- for e?
Should we assume that you know that
##\ln(A)=B##​
is equivalent to
##A=e^B##​
?
 
  • #4
SammyS said:
Should we assume that you know that
##\ln(A)=B##​
is equivalent to
##A=e^B##​
?

Yes I think I understand that part...

This means that e^B > 0, am I right?
 
  • #5
Yes, If we take A to be a function, the range of A valid for log(A) is only positive, but in fact the full range of A is both positive and negative. So as a function itself, A can take both positive and negative values. Hence the exponent is valid in both positive and negative values.
 
  • #6
Alloymouse said:
Yes, If we take A to be a function, the range of A valid for log(A) is only positive, but in fact the full range of A is both positive and negative. So as a function itself, A can take both positive and negative values. Hence the exponent is valid in both positive and negative values.

Ohhh I see. Well I already assumed that. I guess my problem here was with the authors notation. That makes sense, as later on he ended up dropping the +- and just opted for + instead..
 
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  • #7
Rijad Hadzic said:
Yes I think I understand that part...

This means that e^B > 0, am I right?
The fact that B > 0 has little to do with your question.

If you change ##\displaystyle \ \ln \left| \frac {y-2} {y+2} \right| = 4x + c_2 \ ## to an exponential equation, you get:

##\displaystyle \ \left| \frac {y-2} {y+2} \right| = e^{4x + c_2} \ ##​
.
 
Last edited:
  • #8
SammyS said:
The fact that B > 0 has little to do with your question.

If you change ##\displaystyle \ \ln \left| \frac {y-2} {y+2} \right| = 4x + c_2 \ ## to an exponential equation, you get:

##\displaystyle \ \left| \frac {y-2} {y+2} \right| = e^{4x + c_2} \ ##​
.

I see. So because [itex] |\frac {y-2} {y+2} |[/itex] = [itex] \frac {y-2}{y+2} [/itex] for [itex] \frac {y-2} {y+2} >= 0 [/itex]

you get

[itex] \frac {y-2} {y+2} = e^{4x+c_2} [/itex]

and because

[itex] |\frac {y-2} {y+2} |[/itex] = [itex] - \frac {y-2}{y+2} [/itex] for [itex] \frac {y-2} {y+2} < 0 [/itex]

you get

[itex] \frac {y-2} {y+2} = -e^{4x+c_2} [/itex]

so it ends up being two equations?
 
  • #9
Rijad Hadzic said:

Homework Statement


Going from

[itex] ln|\frac {y-2} {y+2}| = 4x + c_2[/itex]

to

[itex] \frac {y-2} {y+2} = \pm e^{4x+c_2} [/itex]

Meaning, why is is [itex] \pm [/itex] e^{4x+c_2}?

Because
$$\ln \left| \frac{y-2}{y+2} \right| = 4x + c \; \Rightarrow \: \left| \frac{y-2}{y+2} \right| = e^{4x+c}$$
and ##A = \pm |A|##.
 
  • #10
Ray Vickson said:
Because
$$\ln \left| \frac{y-2}{y+2} \right| = 4x + c \; \Rightarrow \: \left| \frac{y-2}{y+2} \right| = e^{4x+c}$$
and ##A = \pm |A|##.

Was my reasoning above your post correct?? As to why the equation e is +- and that its really two equatiosn not one
 
  • #11
Rijad Hadzic said:
Was my reasoning above your post correct?? As to why the equation e is +- and that its really two equatiosn not one

Yes, it is correct, and makes my post redundant. However, your post did not appear on my screen until after I posted mine!
 
  • #12
Ray Vickson said:
Yes, it is correct, and makes my post redundant. However, your post did not appear on my screen until after I posted mine!

Haha its all good. My insight is small right now. Even a rewording of something that I'm having trouble with is valuable to me.
 
  • #13
Anytime you have an equation |x|=y, there are two solutions, namely y=x and y=-x. Maybe you already knew that. So, yes there are two equations.
 

Why is this property true?

1. What is the evidence that supports this property?

There are several types of evidence that can support a property, such as experimental data, observations, and mathematical proofs. Scientists use a combination of these to establish the validity of a property.

Why is this property true?

2. Can this property be proven to be true or false?

In science, nothing is ever considered to be absolutely proven. Properties and theories can be supported by evidence, but they can always be challenged or revised as new evidence is discovered.

Why is this property true?

3. What is the underlying mechanism behind this property?

Understanding the mechanism behind a property can provide insight into how and why it occurs. Scientists use theories and models to explain the underlying mechanisms of properties.

Why is this property true?

4. How does this property relate to other properties or theories?

In science, properties and theories are often interconnected. By understanding how a property relates to others, scientists can gain a deeper understanding of the natural world and make predictions about future discoveries.

Why is this property true?

5. How does this property contribute to our understanding of the natural world?

Properties are essential for understanding the natural world. They help us make sense of the complex systems and processes that exist in our universe. By studying properties, we can gain a deeper understanding of the world around us.

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