# Thread: Exp of Log simplification

1. ## Exp of Log simplification

Nice to find this forum!

Could someone help here?

I've this function:

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

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

Really like:

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

Is that possible? Any suggestion?

Thanks!

2. Originally Posted by mamboking
Nice to find this forum!

Could someone help here?

I've this function:

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

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

Really like:

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

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

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

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

3. Originally Posted by mamboking
Nice to find this forum!

Could someone help here?

I've this function:

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

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

Really like:

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

Is that possible? Any suggestion?

Thanks!
I'm not sure how to interpret this part:

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

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