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_{6}(x + 8) + \log_{6}(x + 3) = 1
rewrite the given equation without logarithms. do not solve for x
solve the equation. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the solution set is \boxed{}. (simplify your answer. use a comma to separate answers as needed.)
b. there are infinitely many solutions.
c. there is no solution.

Explanation:

Step1: Apply log product rule

Using the property \(\log_b M + \log_b N=\log_b(MN)\), we get \(\log_6[(x + 8)(x + 3)] = 1\).

Step2: Convert to exponential form

Recall that \(\log_b a = c\) is equivalent to \(b^c=a\). So, \(6^1=(x + 8)(x + 3)\).

Step3: Expand and simplify

Expand the right - hand side: \(x^{2}+3x + 8x+24 = 6\).
Combine like terms: \(x^{2}+11x+24 - 6=0\), which simplifies to \(x^{2}+11x + 18 = 0\).

Step4: Factor the quadratic equation

Factor \(x^{2}+11x + 18\) as \((x + 2)(x+9)=0\).

Step5: Solve for x

Set each factor equal to zero: \(x+2 = 0\) or \(x + 9=0\). So, \(x=-2\) or \(x=-9\).

Step6: Check the domain

For \(\log_6(x + 8)\) and \(\log_6(x + 3)\) to be defined, we need \(x+8>0\) (i.e., \(x>-8\)) and \(x + 3>0\) (i.e., \(x>-3\)).

• When \(x=-9\), \(x+8=-1<0\) and \(x + 3=-6<0\), so \(x=-9\) is not in the domain.
• When \(x=-2\), \(x + 8=6>0\) and \(x+3 = 1>0\), so \(x=-2\) is in the domain.

Answer:

The solution set is \(-2\)