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