Exponential Theory: Explaining e^(-2 ln |x+1|) = e^ln [1/(x+1)^2]

[naspek]

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..

 

[CompuChip]

Nice how you absolutely did not try to hide the fact that you copied this from another forum.

The answer to your question: it is a known calculation rule for logarithms that
y ^alog(x) = ^alog(x^y)
for any number a.

 

[Mark44]

Nice how you absolutely did not try to hide the fact that you copied this from another forum.

The answer to your question: it is a known calculation rule for logarithms that
y ^alog(x) = ^alog(x^y)
for any number a.

I haven't see notation like that before. Is ^alog 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)

 

[naspek]

Thanks Mark44 =)

 

[CompuChip]

I haven't see notation like that before. Is ^alog 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 ^alog is standard notation.
But that is what I meant.

 

[Mark44]

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