Determing range of values of x & y

**moolimanj** · February 3rd 2008, 04:58 AM

Hi

If real variables x and y are related by the equation:

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

then how do i:

a). determine the range of values of x and y for which expressions on each side of equation are defined, and

b). find y explicitly as a function of x, that is, express the equation in the form y=f(x)

Thanks

**earboth** · February 3rd 2008, 05:18 AM

Originally Posted by moolimanj

Hi

If real variables x and y are related by the equation:

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

then how do i:

a). determine the range of values of x and y for which expressions on each side of equation are defined, and

b). find y explicitly as a function of x, that is, express the equation in the form y=f(x)

Thanks

to a) The ln-function is defined for positive arguments:

$3-y>0~\implies~y<3$ and

$2x+1>0~\implies~x>-\frac12$

To b)

$\ln(3-y) = 3\ln(2x+1)-x~\implies~3-y=\frac{(2x+1)^3}{e^x}~\implies~y=3- \frac{(2x+1)^3}{e^x}~,~x>-\frac12$

**red_dog** · February 3rd 2008, 05:23 AM

a) The conditions are
$3-y>0$
$2x+1>0$
Now, can you solve them?

b) The equation can be written as
$\displaystyle\ln(3-y)=\ln(2x+1)^3-\ln e^x\Leftrightarrow\ln(3-y)=\ln\frac{(2x+1)^3}{e^x}\Leftrightarrow$
$\displaystyle\Leftrightarrow 3-y=\frac{(2x+1)^3}{e^x}\Leftrightarrow y=3-\frac{(2x+1)^3}{e^x}$

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