Verified answer

wow I did this like 2 weeks ago and got a 99 on the exam, and I totally forgot how to do these hahah thats sweet.

Log(base 3) (x^2-9) - Log(base 3) (x+1) = 2

log(base3) ((x^2-9) / (x+1) = 2

((x^2-9) / (x+1) = 3^2

x^2-9 = 9(x + 1)

x^2-9 - 9x - 9 = 0

x^2 - 9x - 18 = 0

x = (9 +- sqrt(81 + 72)) / 2

= (9 +- sqrt(153)) / 2

although the negative answer is no good because we end up with a negative numberin the log and you can't take the log of a negative so the answer is

x = (9 + sqrt(153)) / 2

= 10.68 (2dp)

Log(base 3) (x^2-9) - Log(base 3) (x+1) = 2

log(base3)(x^2-9) (divided by) log(base3)(x+1)=2

log(base3) (x^2-9)/(x+1) =2

3^2 = (x^2-9)/(x+1)

9 = (x^2-9)/(x+1)

9(x+1) = (x^2-9)

9x+9 = x^2-9

x^2-9-9x-9 = 0

x^2-9x-18 = 0

then use the quadratic formula from there b/c the roots aren't whole integers

HOPE THIS HELPS N GIVES U COMPREHENSION ^_^

(x^2-9)/(x+1)=9

x^2-9x-18=0

x=10.7