1. ## logarithmics

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!

2. Originally Posted by jvignacio
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

3. just wondering why u made the last line start off with equaling to 1 when you took the log off to find its y value using quadratics and when you have the log base 10 and equation you had it equal to 0.

thanks

4. Originally Posted by jvignacio
just wondering why u made the last line start off with equaling to 1 when you took the log off to find its y value using quadratics and when you have the log base 10 and equation you had it equal to 0.

thanks
if $\displaystyle \log_{10}(u(x))=0\Rightarrow{u(x)=10^0=1}$