(c) $\log _{2} x+\log _{2}(x+2)=3$

Step 1 ：Given the equation \(\log _{2} x+\log _{2}(x+2)=3\)

Step 2 ：Using the property of logarithms that the sum of two logarithms with the same base is the logarithm of the product of the numbers, we can rewrite the equation as \(\log _{2} (x*(x+2))=3\)

Step 3 ：Converting the logarithmic equation to an exponential equation, we get \(2^3 = x*(x+2)\)

Step 4 ：Solving this quadratic equation, we get two solutions: x = -4 and x = 2

Step 5 ：However, we need to check these solutions in the original equation because the logarithm is not defined for negative numbers and zero

Step 6 ：Substituting x = -4 into the original equation, we get \(\log _{2} (-4)+\log _{2}(-4+2)\), which is undefined because the logarithm of a negative number is undefined

Step 7 ：Substituting x = 2 into the original equation, we get \(\log _{2} (2)+\log _{2}(2+2)\), which is equal to 3

Step 8 ：So, the only solution to the equation is x = 2

Step 9 ：Final Answer: \(\boxed{2}\)