1. ## logs

Hey peeps, could somebody help me out with these logs

Solve x for each of the following equations

$log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)$

$log_{3}(x-4)+log_{3}(2x+1)=1$

2. So $\log_{3} \frac{x-2}{2x-2} = \log_{3} 6$. Then $6 = \frac{x-2}{2x-2}$ or $12x-12 = x-2$. Then $11x = 10 \implies x = \frac{10}{11}$.

$\log_{3} \frac{x-4}{2x+1} = 1$ or $3 = \frac{x-4}{2x+1}$. Then $6x+3 = x-4$ and $x = -\frac{7}{5}$.

3. Wow, you're the greatest dude.

4. Originally Posted by gary223
Hey peeps, could somebody help me out with these logs

Solve x for each of the following equations

1. $log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)$

2. $log_{3}(x-4)+log_{3}(2x+1)=1$
Hello,

to #1:

$log_{3}(x+2)-log_{3}(2x-2)=log_{3}6)~\Longrightarrow~ log_{3}\left(\frac{x+2}{2x-2}\right) = log_{3}(6)$ $\Longrightarrow~x+2=12x-12~\Longrightarrow~14 = 11x~\Longrightarrow ~x=\frac{14}{11}$

to #2:

$log_{3}(x-4)+log_{3}(2x+1)=1~\Longrightarrow~ log_{3}\left((x-4)(2x+1)\right)=1$ $\Longrightarrow~2x^2-7x-4=3~\Longrightarrow~2x^2-7x-7=0$ $\Longrightarrow~x = \frac{7}{4}-\frac{1}{4}\cdot \sqrt{105}~\vee~ x = \frac{7}{4}+\frac{1}{4}\cdot \sqrt{105}$

The negative solution doesn't belong to the domain(?) of the equation.