Given the parent functions f (x) = log2 (3x - 9) and g (x) = log2 (x - 3), what is f (x) - g (x) ?. A. f (x) - g (x) = log2 (2x - 6). B. f (x) - g (x) = log2 (2x - 12). C. f (x) - g (x) = log2 one third. D. f (x) - g (x) = log2 3

We are given the two functions:

f (x) = log2 (3x - 9)

g (x) = log2 (x - 3)

Then,

f (x) - g (x) = log2 (3x - 9) - log2 (x - 6)

Two logarithms of the same bases are subtracted, therefore:

log2 ((3x - 9) / (x - 3))

We factor both the numerator and the denominator by x - 3. This becomes,

log2 ((3) (x - 3) / (x - 3))

When further simplified, yields:

log2 (3)

F (x) - g (x) = log2 (3x - 9) - log2 (x - 6)

When two logarithms of the same bases are subtracted they become,

log2 ((3x - 9) / (x - 3))

Both the numerator and the denominator can be factored by x - 3. This becomes,

log2 ((3) (x - 3) / (x - 3))

which simplifies into,

log2 (3).