Solve: log (base 3) (x+5) - log (base 3) (x-3) = 2?

log_3(x + 5) - log_3(x - 3) = 2
log_3[(x + 5)/(x - 3)] = 2
(x + 5)/(x - 3) = 3^2
x + 5 = 9(x - 3)
x + 5 = 9x - 27
x - 9x = -5 - 27
-8x = -32
x = -32/(-8)
x = 4

Ok ILove---- when you distributed the (x+3)(x+5) you should have gotten x^2 + 8x+15. Then subtract 3 from both sides to get x^2 + 8x+12 = 0 and this factors to (x+6)(x+2) = 0 and you have answer x = -6 and -2 but the -6 can't be a valid answer otherwise you have a log of a negative number which is not possible. Looks like you were headed in the right direction with the logs and exponents but simple math error.

let log be log to base 3

log ( x + 5 ) - log ( x - 3 ) = 2

log [ ( x + 5 ) / ( x - 3 ) ] = 2

( x + 5 ) / ( x - 3 ) = 9

x + 5 = 9 x - 27

32 = 8 x

x = 4

then you get log (base 3) [(x+5)/(x-3)] = 2.

Then you rewrite it in the exponential form:

(x+5)/(x-3) = 3^2 = 9

Multiply both sides with (x-3).

x+5 = 3x - 27

x= 16.


**EDIT: Retarded mistake. The answerer below me is right.