# Help with transposition

• May 13th 2009, 01:10 PM
andyw
Help with transposition
Hello,

I'm having particular problems with re-arranging this formulae. (Headbang)
If someone could help me with it i'd be most thankful. Here goes:

X = y/z ln(1+r^3/1- r^3) Make r the subject.

Thanks.
• May 13th 2009, 02:36 PM
Soroban
Hello, Andy!

This is The Transposition Problem from Hell
. . coming soon to a theater near you.

Quote:

Solve for $r\!:\quad x \;=\;\frac{y}{z}\ln\left(\frac{1+r^3}{1-r^3}\right)$

We have: . $\frac{y}{z}\ln\left(\frac{1+r^3}{1-r^3}\right) \;=\;x$

Multiply by $\frac{z}{y}\!:\;\; \ln\left(\frac{1+r^3}{1-r^3}\right) \;=\;\frac{xz}{y}$

Exponentiate: . $\frac{1+r^3}{1-r^3} \;=\;e^{\frac{xz}{y}}$

Multiply by $(1-r^3)\!:\;\;1 + r^3 \;=\;e^{\frac{xz}{y}}(1 - r^3)$

Expand: . $1 + r^3 \;=\;e^{\frac{xz}{y}} - e^{\frac{xz}{y}}r^3$

Rearrange terms: . $e^{\frac{xz}{y}}r^3 + r^3 \;=\;e^{\frac{xz}{y}} - 1$

Factor: . $\left(e^{\frac{xz}{y}} + 1\right)r^3 \;=\;e^{\frac{xz}{y}} - 1$

Then: . $r^3 \;=\;\frac{e^{\frac{xz}{y}} - 1}{e^{\frac{xz}{y}} + 1}$

Therefore: . $r \;=\;\sqrt[3]{\frac{e^{\frac{xz}{y}}-1}{e^{\frac{xz}{t}} + 1}}$

• May 14th 2009, 01:30 AM
andyw
It's amazing how simple it looks once you have the answer!