# help with transpose plz

• Apr 1st 2010, 06:53 AM
matlondon
help with transpose plz
http://img379.imageshack.us/img379/6489/helpmaths.png

how do i get fron u to x, and any good site with transpose of powers.
• Apr 1st 2010, 08:44 AM
tonio
Quote:

Originally Posted by matlondon
http://img379.imageshack.us/img379/6489/helpmaths.png

how do i get fron u to x, and any good site with transpose of powers.

You wrote $u=\left(\frac{2p_2x_2}{p_1}\right)^2\!\!x_2=\left( \frac{2p_2}{p_1}\right)^2\!\!x_2^2x_2=\left(\frac{ 2p_2}{p_1}\right)^2\!\!x_2^3\Longrightarrow u^{1\slash 3}=\left(\frac{2p_2}{p_1}\right)^{2\slash 3}\!\!x_2$ ...and from here isolate $x_2$ and get the result.

Tonio
• Apr 1st 2010, 10:14 AM
Soroban
Hello, matlondon!

Quote:

$U \;=\;\left(\frac{2P_2x_2}{P_1}\right)^2x_2$

$x_2 \;=\;U^{\frac{1}{3}}\left(\frac{P_1}{2P_2}\right)^ {\frac{2}{3}}$

How do i get fron $U$ to $x_2$ ?

We have: . $\left(\frac{2P_1x_2}{P_1}\right)^2x_2 \;=\;U$

. . . . . $\left(\frac{2P_2}{P_1}\right)^2(x_2)^2 \cdot x_2 \;=\;U$

. . . . . . . $\left(\frac{2P_2}{P_1}\right)^2(x_2)^3 \;=\;U$

. . . . . . . . . . . . $(x_2)^3 \;=\;U\left(\frac{P_1}{2P_2}\right)^2$

. . . . . . . . . . . . . . $x_2 \;=\;\left[U\left(\frac{P_1}{2P_2}\right)^2\right]^{\frac{1}{3}}$

. . . . . . . . . . . . . . $x_2 \;=\;U^{\frac{1}{3}}\left(\frac{P_1}{2P_2}\right)^ {\frac{2}{3}}$

• Apr 1st 2010, 03:15 PM
matlondon
yes, but why has he fliped it, p1/ 2p2 or is that a misprint?thx
• Apr 1st 2010, 06:37 PM
tonio
Quote:

Originally Posted by matlondon
yes, but why has he fliped it, p1/ 2p2 or is that a misprint?thx

Because that's what has to be done when passing to the other side of an equation something that was multiplying: if $A\neq 0\,\,\,and\,\,\,Ax=Y\,\,\,then\,\,\,x=\frac{Y}{A}$ . If A happens to be a fraction then when it passes to the other side dividing the fraction "flips". It's simply fractions arithmetic.

Tonio
• Apr 2nd 2010, 03:10 AM
matlondon
Thank you all for helping me, i have another question which i shall post up later.