Hey peeps, could somebody help me out with these logs
Solve x for each of the following equations
$\displaystyle log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)$
$\displaystyle log_{3}(x-4)+log_{3}(2x+1)=1$
So $\displaystyle \log_{3} \frac{x-2}{2x-2} = \log_{3} 6 $. Then $\displaystyle 6 = \frac{x-2}{2x-2} $ or $\displaystyle 12x-12 = x-2 $. Then $\displaystyle 11x = 10 \implies x = \frac{10}{11} $.
$\displaystyle \log_{3} \frac{x-4}{2x+1} = 1 $ or $\displaystyle 3 = \frac{x-4}{2x+1} $. Then $\displaystyle 6x+3 = x-4 $ and $\displaystyle x = -\frac{7}{5} $.
Hello,
to #1:
$\displaystyle log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)~\Longrightarrow~ log_{3}\left(\frac{x+2}{2x-2}\right) = log_{3}(6)$ $\displaystyle \Longrightarrow~x+2=12x-12~\Longrightarrow~14 = 11x~\Longrightarrow ~x=\frac{14}{11}$
to #2:
$\displaystyle log_{3}(x-4)+log_{3}(2x+1)=1~\Longrightarrow~ log_{3}\left((x-4)(2x+1)\right)=1$ $\displaystyle \Longrightarrow~2x^2-7x-4=3~\Longrightarrow~2x^2-7x-7=0$ $\displaystyle \Longrightarrow~x = \frac{7}{4}-\frac{1}{4}\cdot \sqrt{105}~\vee~ x = \frac{7}{4}+\frac{1}{4}\cdot \sqrt{105}$
The negative solution doesn't belong to the domain(?) of the equation.