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solve the logarithmic equation. be sure to reject any value of x that i…

Question

solve the logarithmic equation. be sure to reject any value of x that is not in the domain of the original logarithmic expressions. give the exact answer. \\(\log_{2}(x + 6) + \log_{2}(x + 5) = 1\\) solve the equation. select the correct choice below and, if necessary, fill in the answer box to complete your choice \\(\bigcirc\\) a. the solution set is \\(\boxed{}\\). (simplify your answer. use a comma to separate answers as needed) \\(\bigcirc\\) b. there are infinitely many solutions. \\(\bigcirc\\) c. there is no solution

Explanation:

Step1: Apply log addition rule

Using the property \(\log_b M + \log_b N=\log_b(MN)\), we combine the logs:
\(\log_2[(x + 6)(x + 5)] = 1\)

Step2: Convert to exponential form

Recall \( \log_b a = c \implies b^c = a \). For base \(2\), this gives:
\(2^1=(x + 6)(x + 5)\)
Simplify: \(x^2 + 11x + 30 = 2\)
\(x^2 + 11x + 28 = 0\)

Step3: Solve the quadratic equation

Factor: \((x + 4)(x + 7)=0\)
Solutions: \(x=-4\) or \(x=-7\)

Step4: Check domain of logs

For \(\log_2(x + 6)\) and \(\log_2(x + 5)\), arguments must be \(>0\):

  • \(x=-4\): \(x + 6 = 2>0\), \(x + 5 = 1>0\) (valid)
  • \(x=-7\): \(x + 6=-1<0\), \(x + 5=-2<0\) (invalid, reject)

Answer:

\(-4\)