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

[Rijad Hadzic]

Homework Statement

Going from

ln|\frac {y-2} {y+2}| = 4x + c_2

to

\frac {y-2} {y+2} = \pm e^{4x+c_2}

Meaning, why is is \pm 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?

 

[Alloymouse]

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.

 

[SammyS]

Homework Statement

Going from

ln|\frac {y-2} {y+2}| = 4x + c_2

to

\frac {y-2} {y+2} = \pm e^{4x+c_2}

Meaning, why is is \pm 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

$A=e^B$​

?

 

[Rijad Hadzic]

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?

 

[Alloymouse]

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.

 

[Rijad Hadzic]

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..

 

[SammyS]

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} \$​

.

 

[Rijad Hadzic]

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 |\frac {y-2} {y+2} | = \frac {y-2}{y+2} for \frac {y-2} {y+2} >= 0

you get

\frac {y-2} {y+2} = e^{4x+c_2}

and because

|\frac {y-2} {y+2} | = - \frac {y-2}{y+2} for \frac {y-2} {y+2} < 0

you get

\frac {y-2} {y+2} = -e^{4x+c_2}

so it ends up being two equations?

 

[Ray Vickson]

Homework Statement

Going from

ln|\frac {y-2} {y+2}| = 4x + c_2

to

\frac {y-2} {y+2} = \pm e^{4x+c_2}

Meaning, why is is \pm 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 $A = \pm |A|$.

 

[Rijad Hadzic]

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

 

[Ray Vickson]

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!

 

[Rijad Hadzic]

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.

 

[Gene Naden]

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.