We'll replace a and b by the given logarithms:

log2(x-7) = 6 - log2(3x-5)

We'll add log2(3x-5) both sides:

log2(x-7) + log2(3x-5) = 6

We notice that the bases of logarithms from the left side are equal, therefore, we'll transform the sum into a product:

log2 (x-7)(3x-5) = 6 <=> (x-7)(3x-5) = 2^6

We'll eliminate the brackets using FOIL method:

3x^2 - 5x - 21x + 35 = 64

We'll combine like terms:

3x^2 - 26x + 35 = 64

We'll subtract 64 both sides:

3x^2 - 26x + 35 - 64 = 0

3x^2 - 26x - 29 = 0

We'll apply quadratic formula:

x1 = (26+32)/6

x1 = 58/6 => x1 = 9.(6)

x2 = -1

We'll impose constraints of existence of the logarithms:

x - 7 > 0

x > 7

3x - 5 > 0

3x > 5

x > 5/3

The common interval of admissible values for x is (7 ; +`oo` ).

**We notice that only one value belongs to this interval, therfore the equation will have only one solution, namely x = 9.(6).**