To solve the equation \(\log_2 (x-1) + \log_2 (x-3) = 3\), apply the logarithm property \(\log_b a + \log_b c = \log_b (a \cdot c)\) to combine the terms on the left side:
\(\log_2 ((x-1)(x-3)) = 3\)
Convert the logarithmic equation to an exponential form: if \(\log_2 y = 3\), then \(y = 2^3\).
\((x-1)(x-3) = 2^3\)
\((x-1)(x-3) = 8\)
Expand the left side of the equation:
\(x^2 - 3x - x + 3 = 8\)
\(x^2 - 4x + 3 = 8\)
Rearrange the equation into a standard quadratic form by subtracting 8 from both sides:
\(x^2 - 4x + 3 - 8 = 0\)
\(x^2 - 4x - 5 = 0\)
Factor the quadratic equation. Find two numbers that multiply to -5 and add to -4. These numbers are -5 and 1.
\((x-5)(x+1) = 0\)
Set each factor equal to zero to find potential solutions:
\(x-5=0\) or \(x+1=0\)
This yields:
\(x=5\) or \(x=-1\)
Check the validity of the solutions. Logarithms are defined only for positive arguments. Therefore, \(x-1>0\) and \(x-3>0\), which implies \(x>3\). The solution \(x = -1\) does not satisfy this condition and is extraneous. The only valid solution is \(x = 5\).