$\displaystyle

\ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C

$

can someone rewrite this in terms of y?

I know you need to multiply by e, but I don't know what to do from there.

Thanks.

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- Dec 7th 2008, 10:08 AMwoohoosimplify this function
$\displaystyle

\ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C

$

can someone rewrite this in terms of y?

I know you need to multiply by e, but I don't know what to do from there.

Thanks. - Dec 7th 2008, 10:35 AMSkalkaz
Dear woohoo,

not multiplying.

Tipp: if $\displaystyle ln(y) = C$ than $\displaystyle y = e^C$ - Dec 7th 2008, 10:46 AMwoohoo
K so I put it to the power of e then.

But I'm not quite sure how to simplify it after that (ie. with the + and the - and how to simplify into x and division) - Dec 7th 2008, 10:55 AMMoo
Hello,

It's not "multiplying", but composing with the exponential function.

$\displaystyle \exp(x)=e^x$

$\displaystyle \frac{1}{11} ~ \ln y=\frac{x^2}{2}-\frac{11}{2} ~ \ln(x^2+11)+\frac{C}{11}$

$\displaystyle \ln \left(y^{1/11}\right)=\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}$

$\displaystyle y^{1/11}=\exp \left(\frac{x^2}{2}-\ln \left(\left(x^2+11\right)^{11/2}\right)+\frac{C}{11}\right)$

$\displaystyle y^{1/11}=\frac{\exp \left(\frac{x^2}{2}+\frac{C}{11}\right)}{\exp\left (\ln \left(\left(x^2+11\right)^{11/2}\right)\right)}$

$\displaystyle y^{1/11}=\frac{\exp \left(\frac{x^2}{2}\right) \cdot \exp \left(\frac{C}{11}\right)}{(x^2+11)^{11/2}}$

And honestly, I don't think it is possible to write this in terms of y (Worried) - Dec 7th 2008, 10:55 AMSkalkaz
Note:

$\displaystyle e^{a+b} = e^a*e^b$

But I don't believe you will get much simplier form. - Dec 7th 2008, 11:15 AMwoohoo
- Dec 7th 2008, 11:21 AMwoohoo
*Forgot to mention this, but C is an arbitrary constant.

ie 3C=C

It's value doesn't matter. - Dec 7th 2008, 12:32 PMchabmgph
It does look like there not too much you can do to simplify this, except the piece with ln.

$\displaystyle \ln y \;=\;11\bigg[\frac{x^2}{2} - \frac{11}{2}\ln(x^2+11)\bigg] + C

$

$\displaystyle y = e^{\frac{11}{2}x^2+C}*e^{\frac{-121}{2}\ln (x^2+11)}$

$\displaystyle y = e^{\frac{11}{2}x^2+C}*e^{\ln (x^2+11)^\frac{-121}{2}}$

$\displaystyle y =\bigg[(x^2+11)^\frac{-121}{2}\bigg]*e^{\frac{11}{2}x^2+C}$

- Dec 7th 2008, 08:01 PMwoohoo
Yeah, that didn't work.

But it makes sense to me.

Any more takers? = /