# Thread: rearrrangement of an equation

1. ## rearrrangement 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

2. Originally Posted by wannberocketscientist
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
Not a bad question:

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

Exponentiate each side:

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

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

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

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

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

3. Originally Posted by wannberocketscientist
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
$ln(3 - y) = 3 \cdot ln(2x + 1) - x$

First, a bit of "cleaning up."

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

Now raise both sides to the base e and recall that $e^{ln(z)} = z$:
$3 - y = e^{ln((2x + 1)^3) - x} = (2x + 1)^3 \cdot e^{-x}$

I'm sure you can finish from here.

-Dan

4. thanks for the quick reply!

5. Originally Posted by wannberocketscientist
thanks for the quick reply!
You are welcome wannberocketscientist!