I need to find the following equation in terms of y=f(x):

ln(3-y)=3ln(2x+1)-x

the bit that is stumpin me it the how to get the y out of the initial set of brackets to start the equation.

Thanks for any help

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- Feb 1st 2008, 09:10 AMwannberocketscientistrearrrangement of an equation
I need to find the following equation in terms of y=f(x):

ln(3-y)=3ln(2x+1)-x

the bit that is stumpin me it the how to get the y out of the initial set of brackets to start the equation.

Thanks for any help - Feb 1st 2008, 09:15 AMcolby2152
Not a bad question: (Star)(Star)

$\displaystyle ln(3-y)=3ln(2x+1)-x$

Exponentiate each side:

$\displaystyle e^{ln(3-y)}=e^{3ln(2x+1)-x}$

$\displaystyle 3-y=e^{3ln(2x+1)}e^{-x}$

$\displaystyle 3-y=e^{ln(2x+1)^3}e^{-x}$

$\displaystyle 3-y=(2x+1)^3e^{-x}$

$\displaystyle y=3-(2x+1)^3e^{-x}$ - Feb 1st 2008, 09:16 AMtopsquark
$\displaystyle ln(3 - y) = 3 \cdot ln(2x + 1) - x$

First, a bit of "cleaning up."

$\displaystyle ln(3 - y) = ln((2x + 1)^3) - x$

Now raise both sides to the base e and recall that $\displaystyle e^{ln(z)} = z$:

$\displaystyle 3 - y = e^{ln((2x + 1)^3) - x} = (2x + 1)^3 \cdot e^{-x}$

I'm sure you can finish from here.

-Dan - Feb 1st 2008, 09:17 AMwannberocketscientist
thanks for the quick reply!

- Feb 1st 2008, 09:31 AMcolby2152