problem with logarithms

**srirahulan** · June 16th 2012, 08:04 PM

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

**richard1234** · June 16th 2012, 08:16 PM

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.

**richard1234** · June 21st 2012, 07:20 AM

Whoops, $x-3y +3 = 1$

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