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

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

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