Given a = log2(x-7) and b = log2(3x-5) solve a=6-b?

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).