Math Help!?

Some general rules:
log(x) + log(y) = log(xy)
a^logₐ(x) = x

log₃(x-4) = 2-log₃(x+4)
log₃(x-4) + log₃(x+4) = 2
log₃((x-4)(x+4)) = 2
log₃(x²-16) = 2
x²-16 = 3²
x² = 25
x = ±√25 = ±5
The argument of a logarithm must be positive, so -5 is an extraneous solution.
x = 5

log₃(x - 4) = 2 - log₃(x + 4)
log₃(x - 4) + log₃(x + 4) = 2log₃(3)
log₃(x - 4)(x + 4) = log₃(3²)
(x - 4)(x + 4) = 3²
x² - 16 = 9
x² = 25
x = 5 or x = -5 [rejected, for log₃(-5 - 4) = log₃(-9) is undefined]

Hence, x = 5

log(base 3)(x - 4) = 2 - log(base 3)(x + 4)
log(base 3)(x - 4) + log(base 3)(x + 4) = 2
log(base 3)((x - 4)(x + 4)) = 2
log(base 3)(x^2 - 16) = 2
3^2 = x^2 - 16
x^2 = 25
x = 5 or x = -5
Discard x = -5 because it leads to imaginary numbers, thus a wrong solution, in the original equation, leaving just
x = 5

Let log be log to base 3
log [ ( x - 4 ) ( x + 4 ) ] = 2
( x - 4 ) ( x + 4 ) = 9
x² = 25
x = 5

log₃(x-4) = 2 - log₃(x+4)

log₃(x - 4) + log₃(x + 4) = 2

log₃[(x - 4)(x + 4)] = 2

log₃[x^2 - 16] = 2

x^2 - 16 = 3^2

x^2 = 16 + 9

x^2 = 25

x = ± 5