Answer:
B {8}
Explanation:
I think the double sets per answer option is a mistake.
and we truly have
log2(x) + log2(x - 7) = 3
A {5}
B {8}
C {-1, 5}
D {-1, 8}
E {}
now, first we simply try common sense :
there is no logarithm for a negative number.
so, A is not possible, as (x - 7) = (5 - 7) = -2 is negative.
C and D are not possible, because they contain -1 (a negative number) making the log2 terms invalid.
looking at B :
log2(8) + log2(8 - 7) = 3
3 + log2(1) = 3
3 + 0 = 3
3 = 3
correct !
and because there is a solution, E is incorrect.
let's try this also formally.
remember :
log(xy) = log(x) + log(y)
that means
log2(x) + log2(x - 7) = 3
log2(x(x - 7)) = 3
let's put both sides to the power of 2 :
2^(log2(x(x - 7))) = 2^3
x(x - 7) = 8
x² - 7x = 8
x² - 7x - 8 = 0
a quadratic equation
ax² + bx + c = 0
has the general solution
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -7
c = -8
x = (7 ± sqrt((-7)² - 4×1×-8))/(2×1) =
= (7 ± sqrt(49 + 32))/2 = (7 ± sqrt(81))/2 =
= (7 ± 9)/2
x1 = (7 + 9)/2 = 16/2 = 8
x2 = (7 - 9)/2 = -2/2 = -1
but as explained above, there is no logarithm for a negative number like -1. and so, the terms in the original equation log2(-1) and log2(-1 - 7) are invalid.
therefore, the only valid solution is
x = 8
and therefore, B is correct.