Bases of logarithms are the same. Let's move the logarithms to the left side and the number to the right side.

log 2 (x+4)+log 2 (x+6)- log2 (x+5)=3

Use the product property for the logarithms:

log 2 (x+4)+log 2 (x+6)=log 2 (x+4)(x+6)

Use the quotient property for the logarithms:

log 2 (x+4)(x+6) - log2 (x+5)= log 2 [(x+4)(x+6)/(x+5)]

Write the equation.

log 2 [(x+4)(x+6)/(x+5)]=3 =>

=> [(x+4)(x+6)/(x+5)]=2^3

[(x+4)(x+6)/(x+5)]=8

(x+4)(x+6) = 8(x+5)

Open the brackets

x^2 + 10x + 24 - 8x - 40 = 0

x^2 + 2x - 16=0

x1=(-2+sqrt68)/2 => x1=-1+sqrt17

x2=-1-sqrt17

Since the values of x for the logarithms to be defined is (-4,+oo), the value -1-sqrt17 is not included.

**ANSWER: The equation ihas a single solution x1=-1+sqrt17.**

`log_2 (x+4) - 3 = log_2 (x+5) - log_2 (x+6)`

`log_2 (x+4) - log_2 2^3 = log_2 (x+5) - log_2 (x+6)`

`log_2 (x+4)/8 = log_2 ((x+5)/(x+6))`

`==> (x+4)/8 = (x+5)/(x+6)`

`==> (x+4)(x+6) = 8(x+5)`

`==> x^2 + 10x + 24 = 8x + 40`

`==> x^2 +2x - 16 = 0`

`==> x1= (-2+sqrt68)/2 = (-2+2sqrt17)/2 = -1+sqrt17`

`==> x2= -1-sqrt17`

But x2 is not a valid answer because the log is not defined.

==> Then the answer is : `x = -1+sqrt17` ``

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