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solve loge (x solve log₂(x - 4) + log₂(x - 10) = 4 what are the solutio…

Question

solve loge (x
solve log₂(x - 4) + log₂(x - 10) = 4
what are the solution sets

Explanation:

Step1: Apply Logarithm Product Rule

Using the property \(\log_a M + \log_a N=\log_a(MN)\), we combine the two logarithms:
\(\log_2[(x - 4)(x - 10)] = 4\)

Step2: Convert to Exponential Form

Recall that \(\log_a b = c\) is equivalent to \(a^c = b\). So for base \(2\), we have:
\(2^4=(x - 4)(x - 10)\)
\(16=x^2-10x - 4x + 40\)
\(x^2-14x + 24 = 0\)

Step3: Solve the Quadratic Equation

Factor the quadratic:
\((x - 12)(x - 2)=0\)
So \(x = 12\) or \(x = 2\)

Step4: Check Domain

The domain of \(\log_2(x - 4)\) and \(\log_2(x - 10)\) requires \(x-4>0\) (i.e., \(x > 4\)) and \(x - 10>0\) (i.e., \(x>10\)).

  • For \(x = 2\), \(2-10=-8<0\), so it's invalid.
  • For \(x = 12\), \(12 - 4 = 8>0\) and \(12 - 10 = 2>0\), so it's valid.

Answer:

\(\{12\}\)