Let's solve the logarithmic equation log base 2 of x minus 3 plus log base 2 of x plus 1 equals 3. First, we need to determine the domain. For logarithms to be defined, their arguments must be positive. So x minus 3 must be greater than 0, which means x is greater than 3. And x plus 1 must be greater than 0, which means x is greater than negative 1. The intersection of these conditions gives us the domain: x must be greater than 3.
Now we'll combine the logarithms using the logarithm property. The property states that log base b of M plus log base b of N equals log base b of M times N. Applying this to our equation, log base 2 of x minus 3 plus log base 2 of x plus 1 becomes log base 2 of the product x minus 3 times x plus 1, which still equals 3. Next, we convert this logarithmic equation to exponential form. Since log base 2 of something equals 3, that something must equal 2 to the power of 3, which is 8.
Now we solve the quadratic equation. First, we expand the left side: x minus 3 times x plus 1 equals x squared plus x minus 3x minus 3, which simplifies to x squared minus 2x minus 3. Setting this equal to 8, we get x squared minus 2x minus 3 equals 8. Moving 8 to the left side gives us the standard form: x squared minus 2x minus 11 equals 0. Using the quadratic formula with a equals 1, b equals negative 2, and c equals negative 11, we get x equals 2 plus or minus the square root of 4 plus 44, all divided by 2.
Now we simplify our solution. First, we simplify the square root of 48. Since 48 equals 16 times 3, the square root of 48 equals 4 times the square root of 3. Substituting this back, we get x equals 2 plus or minus 4 square root of 3, all divided by 2. Simplifying the fraction gives us x equals 1 plus or minus 2 square root of 3. This gives us two potential solutions: x equals 1 plus 2 square root of 3, which is approximately 4.46, and x equals 1 minus 2 square root of 3, which is approximately negative 2.46.
Now we check our solutions against the domain constraint. We need x to be greater than 3. The first solution, 1 plus 2 square root of 3, is approximately 4.46, which is indeed greater than 3, so it's valid. The second solution, 1 minus 2 square root of 3, is approximately negative 2.46, which is less than 3, so it's extraneous and must be rejected. Therefore, our final answer is x equals 1 plus 2 square root of 3. We can verify this by substituting back into the original equation, which confirms our solution is correct.