Posted by **Kenya** on Saturday, April 30, 2011 at 11:08pm.

Solve the equation.

2log3 y = log3 4+ log3 (y+8)

Here's what ive got

take out the logs and say y^2= 4+ y +8

which simplifies to (y^2)-y=12

how do i simplfy the y?

- Algebra -
**Jai**, Sunday, May 1, 2011 at 1:21am
note that you cannot simplify the right side equation log3 (4) + log3 (y+8) , by 4 + y + 8 ... they should be multiplied. recall some laws of exponents; if we are add terms of the same base, for example

log 3 + log 5

this simplifies to

log(3*5) or log 15

going back to the problem, the equation simplifies to

2log3 y = log3 4 + log3 (y+8)

log3 (y^2) = log3 (4*(y+8))

log3 (y^2) = log3 (4y + 32)

cancelling out the log3,

y^2 = 4y + 32

y^2 - 4y - 32 = 0

this is a quadratic equation. since it's factorable, let's just factor this one:

(y - 8)(y + 4) = 0

y = 8 and y = -4

note that y = -4 is extraneous, since if we substitute this back to the original, and take its log:

left side: 2log3 (-4) = undefined

thus,

y = 8

hope this helps~ :)

- Algebra -
**Anonymous**, Sunday, May 1, 2011 at 10:45am
Jai, your a life saver! you explained everything perfectly! thank you sooo much!

that helped a ton, thanks! : )

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