# Find b value if log(b+2)^2-2logb=log3.Find b value if log(b+2)^2-2logb=log3.

*print*Print*list*Cite

### 1 Answer

We'll establish the conditions of existence of logarithms:

(b+2)^2>0

The perfect square is always positive, for any value of b.

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

We'll shift 2log b to the right:

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 we have matching bases both sides, 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 it's not important if b is positive or negative, we'll accept both values of b as solutions for the given equation: {1 - sqrt 3 ; 1 + sqrt 3}.