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e ^ (-2 ln |x + 1|) = e ^ ln [1 / (x + 1)^2]

how can this happen?

can anyone explain to me the process of this equation..

how can this happen?

can anyone explain to me the process of this equation..

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- Thread starter naspek
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how can this happen?

can anyone explain to me the process of this equation..

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- #2

CompuChip

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The answer to your question: it is a known calculation rule for logarithms that

y

for any number

- #3

Mark44

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The answer to your question: it is a known calculation rule for logarithms that

y^{a}log(x) =^{a}log(x^{y})

for any numbera.

I haven't see notation like that before. Is

The notation that is used more often for this property of logarithms, I believe, is this:

log

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Thanks Mark44 =)

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CompuChip

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I haven't see notation like that before. Is^{a}log supposed to represent the log base a of something?

The notation that is used more often for this property of logarithms, I believe, is this:

log_{a}(x^{y}) = y log_{a}(x)

Right, sorry.

Where I come from

But that is what I meant.

- #6

Mark44

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I figured that's what it meant, but it's something I haven't run across it before.

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