Can't find the determinant of the Jacobian

[Addez123]

Homework Statement

Determin the determinant for the image (x,y) -> (u,v)
u = a^2 + b^2
v = a^2 - b^2
a = ln(xy)
b = ln(x/y)

Relevant Equations

Matrices

The way I approach it was, we're looking for det(H) where H = h(u, v)
$$H = \begin{bmatrix}
du/da & du/db \\
dv/da & dv/db
\end{bmatrix} *
\begin{bmatrix}
da/dx & da/dy \\
db/dx & db/dy
\end{bmatrix}$$

I just multiply those two matrices and then get the determinant. The answer is
$$16((ln x)^2 - (ln y)^2)/(xy)$$

But all I get is
$$16*ln(xy) * ln(x/y)/(xy)$$

I've counted it twice so if something is wrong its the matrices or the way I think.

 

[PeroK]

You have the same function for $u$ and $v$.

 

[Addez123]

You have the same function for $u$ and $v$.

sorry! v was suppose to have a minus sign in it.

 

[PeroK]

I just multiply those two matrices and then get the determinant. The answer is
$$16((ln x)^2 - (ln y)^2)/(xy)$$

But all I get is
$$16*ln(xy) * ln(x/y)/(xy)$$

Are they different?

 

[Addez123]

Unless ln(xy) = 2 they are, right?

 

[PeroK]

Unless ln(xy) = 2 they are, right?

If you have a spreadsheet on your computer, why not plug in some values for $x$ and $y$ and see?

 

[PeroK]

And/or, do a bit of algebra.

 

[Addez123]

I plugged both equations into wolfram, they are not the same.

Trying to make my equation look like the first equation results in this

$$16((ln x)^{ln(xy)} - (ln y)^{ln(xy)})/(xy)$$
which is not the same.

 

[PeroK]

I plugged both equations into wolfram, they are not the same.

Trying to make my equation look like the first equation results in this

$$16((ln x)^{ln(xy)} - (ln y)^{ln(xy)})/(xy)$$
which is not the same.

Give me an example of $x, y$ where: $$(ln x)^2 - (ln y)^2 \ne ln(xy)ln(x/y)$$

 

[Orodruin]

Did you try simplifying a bit by using logarithm relations such as ln(xy) = ln(x) ln(y)?

 

[PeroK]

Did you try simplifying a bit by using logarithm relations such as ln(xy) = ln(x) ln(y)?

I hope that one isn't in your book!

 

[Addez123]

Give me an example of $x, y$ where: $$(ln x)^2 - (ln y)^2 \ne ln(xy)ln(x/y)$$

Wow. Turns out it was the same!
Turning ln(xy) = ln(x) + ln(y)
and ln(x/y) = ln(x) - ln(y)
and then just multiplying gives the correct answer.

Wierd I didnt see that!
Thanks!