$\displaystyle log(x-2)+log2=2logy;\ log(x-3y+3)=0$ find out the value of (x) and (y).

Printable View

- Jun 16th 2012, 08:04 PMsrirahulanproblem with logarithms
$\displaystyle log(x-2)+log2=2logy;\ log(x-3y+3)=0$ find out the value of (x) and (y).

- Jun 16th 2012, 08:16 PMrichard1234Re: problem with logarithms
Note that $\displaystyle \log(x-2) + \log 2 = \log(2x-4)$, and that $\displaystyle 2 \log y = \log y^2$. Therefore the first equation becomes

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

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

Solve the system of equations. - Jun 21st 2012, 07:20 AMrichard1234Re: problem with logarithms
Whoops, $\displaystyle x-3y +3 = 1$