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|dw:1358409590944:dw|
\[\log_3(x+4)+\log_3(x-2)=3\]log a + log b = log (ab) \[\log_3(x+4)(x-2)=3\] (x+4)(x-2) = 3^3
so would it be x^2+2x-17=0 which makes the roots x = {3.243, -5.243}?

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Other answers:

Not quite, can you check again?
I'm sorry I can't seem to find where im going wrong
The equation you got was wrong..
(x+4)(x-2) = 3^3 x^2 + 2x - 8 - 3^3 =0 ^Simplify this
is it not x^2+2x-17 ?
what is 3^3?
oh my apologies I was thing of multiplying, so it is x^2+2x+19?
Hmmm... x^2 + 2x - 8 - 3^3 =0
x^2+2x-35=0 ?
Yup :) Now, solve it~
x=5 and x=-7 ?
Yes, but you have to reject one answer there.
You cannot take log for a negative number.
-7 correct?
ok thank you so much
You're welcome :)
And yes, you need to reject -7 :)

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