# problem with logarithms

• June 16th 2012, 09:04 PM
srirahulan
problem with logarithms
$log(x-2)+log2=2logy;\ log(x-3y+3)=0$ find out the value of (x) and (y).
• June 16th 2012, 09:16 PM
richard1234
Re: problem with logarithms
Note that $\log(x-2) + \log 2 = \log(2x-4)$, and that $2 \log y = \log y^2$. Therefore the first equation becomes

$\log(2x-4) = \log y^2 \Rightarrow 2x-4 = y^2$

The second equation, note that $\log 1 = 0$ (for any non-zero base), so $x - 3y + 3 = 0$.

Solve the system of equations.
• June 21st 2012, 08:20 AM
richard1234
Re: problem with logarithms
Whoops, $x-3y +3 = 1$