# Thread: solve for log equation

1. ## solve for log equation

log3x + log3(x + 2) = 1

how do i prove this?

So far, i have:
= log3[x(x+2)]
= log3[x^2 + 2x]

and i can't get any farther?

2. Originally Posted by snypeshow
log3x + log3(x + 2) = 1

how do i prove this?

So far, i have:
= log3[x(x+2)]
= log3[x^2 + 2x]

and i can't get any farther?
Change is from a log to a non log form; therefore, it will now look like this:

$\displaystyle x^2+2x=3^1$

3. so I would get x = sqrt(3 - 2x)

?

4. Yes, because to get rid of the power u need to square root the rest

5. YES! thanks alot guys!!!!

man, there's nothing you guys cant answer :P

6. Originally Posted by snypeshow
so I would get x = sqrt(3 - 2x)

?

$\displaystyle x^2+2x=3^1$

move the subtract 3 from both sides

$\displaystyle x^2+2x-3=0$

factor

$\displaystyle (x+3)(x-1)=0$

solve for x:

x=-3, x=1

check to make sure you do not get a negative log .... since -3 does give you a negative log

7. Originally Posted by RRH

$\displaystyle x^2+2x=3^1$

move the subtract 3 from both sides

$\displaystyle x^2+2x-3=0$

factor

$\displaystyle (x+3)(x-1)=0$

solve for x:

x=-3, x=1

check to make sure you do not get a negative log .... since -3 does give you a negative log