Exp of Log simplification

**mamboking** · May 30th 2008, 08:52 AM

Nice to find this forum!

Could someone help here?

I've this function:

$ f(u,v) = e^{[\ln(v)^{-1/\beta}\ln(u)^{-1/\beta}]^{-\beta}} $

that I need to simplify in a polynomial form with neither
exp nor log but with all the exponents in evidence.

Really like:

$ f(u,v) = [polynom(v)^{-1/\beta}\ ...\ polynom(u)^{-1/\beta}]^{-\beta} $

Is that possible? Any suggestion?

Thanks!

**Mathstud28** · May 30th 2008, 11:48 AM

Originally Posted by mamboking

Nice to find this forum!

Could someone help here?

I've this function:

$ f(u,v) = e^{[\ln(v)^{-1/\beta}\ln(u)^{-1/\beta}]^{-\beta}} $

that I need to simplify in a polynomial form with neither
exp nor log but with all the exponents in evidence.

Really like:

$ f(u,v) = [polynom(v)^{-1/\beta}\ ...\ polynom(u)^{-1/\beta}]^{-\beta} $

Is that possible? Any suggestion?

Thanks!

I dont think this can be done, sorry to say, you could do fairly well except what to do with

$e^{ln(x)^{\frac{-1}{\beta}}}$

which to my knowledge cannot be simplified...I might be missing something simple though

**Chris L T521** · May 30th 2008, 11:59 AM

Originally Posted by mamboking

Nice to find this forum!

Could someone help here?

I've this function:

$ f(u,v) = e^{[\ln(v)^{-1/\beta}\ln(u)^{-1/\beta}]^{-\beta}} $

that I need to simplify in a polynomial form with neither
exp nor log but with all the exponents in evidence.

Really like:

$ f(u,v) = [polynom(v)^{-1/\beta}\ ...\ polynom(u)^{-1/\beta}]^{-\beta} $

Is that possible? Any suggestion?

Thanks!

I'm not sure how to interpret this part:

$\ln(v)^{-\frac{1}{\beta}}$

is it: $\ln(v^{-\frac{1}{\beta}})$ or $(\ln(v))^{-\frac{1}{\beta}}$ ??

Thread: Exp of Log simplification

LinkBack

Thread Tools

Display

Exp of Log simplification

Similar Math Help Forum Discussions

I don't get this simplification

Simplification

Simplification

Simplification

Simplification

Search Tags