Solving the Logarithmic Equation

The problem requires finding the value of x that satisfies the given logarithmic equation:

$ \log x + \log (x - 7) = \log (x + 11) + \log 2 $

Applying Logarithm Properties

Use the logarithm property $ \log a + \log b = \log (ab) $ to simplify both sides of the equation.

• Left side: $ \log (x(x - 7)) $
• Right side: $ \log (2(x + 11)) $

The equation becomes:

$ \log (x(x - 7)) = \log (2(x + 11)) $

Equating Arguments and Solving

Since the logarithms are equal, their arguments must be equal:

$ x(x - 7) = 2(x + 11) $

Expand both sides:

$ x^2 - 7x = 2x + 22 $

Rearrange into a standard quadratic equation ($ ax^2 + bx + c = 0 $):

$ x^2 - 7x - 2x - 22 = 0 $

$ x^2 - 9x - 22 = 0 $

Factor the quadratic equation:

$ (x - 11)(x + 2) = 0 $

This gives two potential solutions: $ x = 11 $ or $ x = -2 $.

Checking Domain Restrictions

The argument of a logarithm must always be positive.

• For $ \log x $, we need $ x > 0 $.
• For $ \log (x - 7) $, we need $ x - 7 > 0 $, which means $ x > 7 $.
• For $ \log (x + 11) $, we need $ x + 11 > 0 $, which means $ x > -11 $.

To satisfy all conditions, x must be greater than 7 ($ x > 7 $).

Validating Potential Solutions

Evaluate the potential solutions against the domain requirement ($ x > 7 $):

• If $ x = 11 $: $ 11 > 7 $. This solution is valid.
• If $ x = -2 $: $ -2 $ is not greater than 7. This solution is extraneous and must be discarded.

Therefore, the only value of x that satisfies the original equation is 11.