# P+q = pq Hölder

## Homework Statement

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])

## Homework Equations

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

## The Attempt at a Solution

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

TIA!

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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.

ehild
Homework Helper
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])

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

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