log (b+2)^2- 2log b = log 3

We will use the algorethim properties to solve:

First we know that: a*log b = log b^a

==> 2log (x+2) - 2log b = log 3

Now factor 2:

==> 2(log(b+2) - log b) = log 3

Now we know that :

log a - log b = log a/b

==> 2 log (b+2)/b) = log 3

==> 2 log (1+ 2/b) = log 3

==> log (1+2/b)^2 = log 3

==> (1+ 2/b)^2 = 3

==> 1 + 4/b + 4/b^2 = 3

==> (b^2 + 4b + 4) = 3b^2

==> 2b^2 - 4b - 4 = 0

Divide by 2:

==> b^2 - 2b -2 = 0

==> b1= (2 + sqrt(4+8)/ 2 = (2+ 2sqrt3)/2 = 1+ sqrt3

==> b2= 1- sqrt3

We'll impose constraints of existence of logarithms:

(b+2)^2>0

The binomial raised to square is always positive, for any value of b.

log (b+2)^2- 2log b = log 3

We'll add 2log b both sides:

log (b+2)^2 = log 3 + 2log b

We'll use the power property of logarithms:

2log b = log b^2

log (b+2)^2 = log 3 + log b^2

We'll use the product rule of logarithms:

loga + log b = log a*b

log 3 + log b^2 = log 3*b^2

log (b+2)^2 = log 3*b^2

Since the bases are matching, we'll apply one to one rule:

(b+2)^2 = 3*b^2

We'll expand the square:

b^2 + 4b + 4 - 3*b^2 = 0

We'll combine like terms:

-2b^2 + 4b + 4 = 0

We'll divide by -2:

b^2 - 2b - 2 = 0

We'll apply the quadratic formula:

b1 = [2 + sqrt(4 + 8)]/2

b1 = (2+2sqrt3)/2

b1 = 1 + sqrt 3

b2 = 1 - sqrt 3

**Since any value of b is accepted, we'll have 2 solutins for the given equation: {1 - sqrt 3 ; 1 + sqrt 3}.**