It'll probably be easier if I use that equation thing to ask this o: Lol.

Question

anonymous · Answer

$$\log_{5} (y^2 +5y + 6) = \log_{5} (y + 3) + \log_{5} 4$$

anonymous · Answer

And whoever figures it out, if you could show me how to do it that'd be great. c:

zarkon · Answer

I guess I'll reply

$$\log_{5} (y^2 +5y + 6) = \log_{5} (y + 3) + \log_{5} (4)$$

$$\log_{5} (y^2 +5y + 6) = \log_{5} [4(y + 3)]$$


$$y^2 +5y + 6=4(y + 3)$$

solve for $y$ then check solution in original equation

anonymous · Answer

@Typically Can you continue?

anonymous · Answer

I.. think so. o: Lul.

anonymous · Answer

Just making sure :)

anonymous · Answer

Hmn, can you show me just in case? I'm practicing for an exam o;

anonymous · Answer

$$y^2 +5y + 6=4(y + 3)$$
multiply out
$$y^2+5y+6=4y+12$$set = 0
$$y^2+y-6=0$$factor
$$(y+3)(y-2)=0$$solve for $y$
$y=-3$ or $y=2$ then check your answer remembering that you cannot take the log of a negative number

zarkon · Answer

or the log of zero :)

anonymous · Answer

So.. the other side doesn't matter after that?