finding all solutions to this question.
$\displaystyle
2\log y = log(3-2y)
$
$\displaystyle
2\log y = log(\frac {3} {2y})
$
$\displaystyle
2\log y = \frac {3} {2y}
$
Now.. im not to sure what to do with the
$\displaystyle
2\log y
$
thank u!
finding all solutions to this question.
$\displaystyle
2\log y = log(3-2y)
$
$\displaystyle
2\log y = log(\frac {3} {2y})
$
$\displaystyle
2\log y = \frac {3} {2y}
$
Now.. im not to sure what to do with the
$\displaystyle
2\log y
$
thank u!
Woah! We have got some problems here
$\displaystyle 2\log_{10}(y)=\log_{10}(y^2)=\log_{10}(3-2y)\Rightarrow\log_{10}(y^2)-\log_{10}(3-2y)=\log_{10}\left(\frac{y^2}{3-2y}\right)=0$
Now exponentiating both sides we get
$\displaystyle \frac{y^2}{3-2y}=1\Rightarrow{y^2-3+2y=0}\Rightarrow{y=\frac{-2\pm\sqrt{4-4(-3)(1)}}{2}}=-1\pm{2}$
One of which is an extraneous solution