# P+q = pq Hölder

1. Aug 11, 2010

### zxh

1. The problem statement, all variables and given/known data

How do you get from 1/p +1/q = 1 to (p-1)(q-1) = 1? (http://de.wikipedia.org/wiki/Konjugation_%28Reelle_Zahlen%29" [Broken])

2. Relevant equations

Is log(xy) = log(x) + log(y) in any way related to this, i.e the rewritten subject title?

3. The attempt at a solution

Manipulate[
ContourPlot[x + y == x*y, {x, -m, m}, {y, -m, m},
PlotRange -> Automatic], {m, -100, 100}]

TIA!

Last edited by a moderator: May 4, 2017
2. Aug 11, 2010

### disregardthat

Multiply out and compare. Logarithms have nothing to do with the equation you mentioned, but you might be using the logarithm to prove some other aspect of Hölders inequality.

3. Aug 11, 2010

### ehild

You can get the second equation from the first one by multiplying both sides of the first equation by pq, then subtracting 1, and factoring. Try, it is fun.

ehild

Last edited by a moderator: May 4, 2017
4. Aug 11, 2010

### zxh

Alright, makes me look tired, but i thought the similarity with the logarithm was too salient not to connect the two.