Is this how you solve logarithmic equations

For log3 (x-5) + log3 (x + 3) = 2 where 3 is the base.

Would you have (x- 5) (x + 3) equal to 3^2 which equals 9, so x would be either 14 or 6?

What about log2 (x - 6) + log2 (x - 4) - log2 x = 2? where 2 is the base. What do you do for that one, the book doesn't have any problems similar to this.

log[3](x - 5) + log[3](x + 3) = 2
log[3]((x - 5) * (x + 3)) = 2
(x - 5) * (x + 3) = 3^2
x^2 - 5x + 3x - 15 = 9
x^2 - 2x - 24 = 0
(x - 6) * (x + 4) = 0
x = -4 , 6

x cannot be -4 because log[3](-4 - 5) and log[3](-4 + 3) do not exist

x = 6



log[2](x - 6) + log[2](x - 4) - log[2](x) = 2
log[2]((x - 6) * (x - 4) / x) = 2
(x - 6) * (x - 4) / x = 2^2
x^2 - 6x - 4x + 24 = 4 * x
x^2 - 6x - 4x - 4x + 24 = 0
x^2 - 14x + 24 = 0
(x - 12) * (x - 2) = 0
x = 2 , 12

x cannot equal 2 because log(2 - 6) and log(2 - 4) do not exist

x = 12

log₃ (x-5) + log₃ (x + 3) = 2
log₃[(x - 5)(x + 3)] = 3^2
(x - 5)(x + 3) = 9
x^2 -2x - 15 = 9
x^2 - 2x - 15 - 9 = 0
x^2 - 2x - 24 = 0
(x + 4)(x - 6) = 0
x + 4 = 0, x - 6 = 0
x = - 4, x = 6

discard x = - 4, since it is a negative number..
the real answer is 6 answer//


log₂(x - 6) + log₂(x - 4) - log₂(x) = 2
.........(x - 6)(x - 4)
log₂[ --------------------] = 2^2
................x

..(x - 6)(x - 4)
---------------------- = 4
........x

x^2 - 10x + 24 = 4x
x^2 - 10x - 4x + 24 = 0
x^2 - 14x + 24 = 0
(x - 12)(x - 2) = 0
x - 12 = 0 , x - 2 = 0
x = 12, x = 2

x= 2 in not an answer of log2 (x - 6) + log2 (x - 4) - log2 x = 2, since when we check it, it doesn't equal to 2..http://www.wolframalpha.com/input/?i=log…

x = 12 answer //

log_3 (x - 5) + log_3 (x + 3) = 2
log_3 [(x - 5)(x + 3)] = 2
(x - 5)(x + 3) = 3^2 = 9
x^2 - 2x - 15 = 9
x^2 - 2x - 24 = 0
(x - 6)(x + 4) = 0
x - 6 = 0 or x + 4 = 0
x = 6............x = -4 (reject)
log_3 (-4) is undefined.

log_2 (x - 6) + log_2 (x - 4) = log_2 x = 2
log_2 [(x - 6)(x - 4)/x] = 2
(x - 6)(x - 4)/x = 2^2 = 4
(x - 6)(x - 4) = 4x
x^2 - 10x + 24 = x
x^2 - 11x + 24= 0
(x - 3)(x - 8) = 0
x - 3 = 0 or x - 8 = 0
x = 3 (reject)........x = 8
3 - 4 = -1 and log_2 (-1) is undefined.

llog3 (x-5) + log3 (x + 3) = 2
Log3[(x-5)*(x+3)] =2
[(x-5)*(x+3)]= 9 ----> x2-2x-24=0 x=-6 x=4

Log2 (x - 6) + log2 (x - 4) - log2 x = 2
Log2[(x-6)*(x-4)/x] =2
[(x-6)*(x-4)/x] = 4 ----> x2-14x+24=0 x=-12 x=-2

solve for x
(x-6)(x-4)/x = 2^2 , then cgheck the solution(s)