Question

homework: practice what you missed final exam part a question solve the logarithmic equation. \\(\log_{5}(x + 3) + \log_{5}(x + 7) = 1\\) select the correct choice below and, if necessary, fill in the answer box to complete your choice. \\(\bigcirc\\) a. \\(x = \square\\) (simplify your answer.) \\(\bigcirc\\) b. there is no solution.

Explanation:

Step1: Apply log addition rule

Using the property \(\log_a M+\log_a N = \log_a(MN)\), we get \(\log_5[(x + 3)(x + 7)]=1\).

Step2: Convert to exponential form

Recall that \(\log_a b=c\) is equivalent to \(a^c = b\). So, \(5^1=(x + 3)(x + 7)\).

Step3: Expand and simplify

Expand the right - hand side: \(x^{2}+7x + 3x+21 = 5\).
Combine like terms: \(x^{2}+10x+21 - 5=0\), which simplifies to \(x^{2}+10x + 16 = 0\).

Step4: Solve the quadratic equation

Factor the quadratic: \((x + 2)(x+8)=0\).
Set each factor equal to zero: \(x+2 = 0\) or \(x + 8=0\). So, \(x=-2\) or \(x=-8\).

Step5: Check domain of logarithm

For \(\log_5(x + 3)\) and \(\log_5(x + 7)\) to be defined, \(x+3>0\) (i.e., \(x>-3\)) and \(x + 7>0\) (i.e., \(x>-7\)).

• For \(x=-2\): \(x+3=-2 + 3=1>0\) and \(x + 7=-2+7 = 5>0\), so \(x=-2\) is valid.
• For \(x=-8\): \(x+3=-8 + 3=-5<0\), so \(x=-8\) is not valid.

Answer:

A. \(x = - 2\)