Given that (a-b)(a+b)=14ab-2b 2

As follows:

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We are given that:

(a-b)(a+b)=14ab-2b^2       expand on the left

a^2 - b^2  = 14ab - 2b^2     add 2b^2  to both sides

a^2 + b^2  = 14ab        divide both sides by 14

ab =  [ a^2 + b^2] / 14      (1)

So, prove that

log₂(a + b)  = 2 + log₂(ab)/2 is true    [let this be (2) ]

log₂(a + b)  = 2 + log₂(ab)/2       multiply both sides by 2

log2(a + b)  = 4 + log2(ab)/2          write 4 as log₂(16)

2log₂(a + b)  = log₂(16) + log₂ [(a^2 + b^2) / 14]

log₂(a + b)^2  = log₂ [ (16/14)(a^2 + b^2) ]     now,  we can forget the logs

(a + b)^2  = ((16/14)(a^2 + b^2)           [16/14  = 8/7]

a^2 + 2ab + b^2  = (8/7)a^2 + (8/7)b^2       subtract a^2, b^2  from both sides

2ab = (1/7)a^2 + (1/7)b^2

2ab  = [a^2 + b^2]/7    divide both sides by 2

ab = [a^2 + b^2] / 14        which equals (1) and which was assumed true, so (2) must be true, as well