To determine the value of [tex]\( x \)[/tex] that satisfies the equation
[tex]\[
\log_6(x + 10) + \log_6(x + 5) = 2,
\][/tex]
we can proceed as follows:
1. Combine the logarithms: Using the logarithm property [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex], we combine the logarithms on the left-hand side:
[tex]\[
\log_6((x + 10)(x + 5)) = 2.
\][/tex]
2. Remove the logarithm: To eliminate the logarithm, we rewrite the equation in exponential form. The equation [tex]\(\log_b(y) = z\)[/tex] is equivalent to [tex]\(b^z = y\)[/tex]. Therefore, we have:
[tex]\[
(x + 10)(x + 5) = 6^2.
\][/tex]
3. Simplify: Calculate [tex]\(6^2\)[/tex] to get 36. So, we rewrite the equation as:
[tex]\[
(x + 10)(x + 5) = 36.
\][/tex]
4. Expand and form a quadratic equation: Expand the left-hand side:
[tex]\[
x^2 + 5x + 10x + 50 = 36,
\][/tex]
which simplifies to:
[tex]\[
x^2 + 15x + 50 = 36.
\][/tex]
5. Set the equation to zero: Subtract 36 from both sides to set the equation to zero:
[tex]\[
x^2 + 15x + 14 = 0.
\][/tex]
6. Solve the quadratic equation: Factor the quadratic equation [tex]\(x^2 + 15x + 14 = 0\)[/tex]:
[tex]\[
(x + 14)(x + 1) = 0.
\][/tex]
This gives us two possible solutions:
[tex]\[
x + 14 = 0 \quad \text{or} \quad x + 1 = 0.
\][/tex]
So,
[tex]\[
x = -14 \quad \text{or} \quad x = -1.
\][/tex]
7. Verify the solutions: We need to check if these solutions are valid by ensuring the arguments of the logarithms are positive:
- For [tex]\(x = -14\)[/tex]: [tex]\(x + 10 = -4\)[/tex] and [tex]\(x + 5 = -9\)[/tex], both of which are negative and not valid for logarithms.
- For [tex]\(x = -1\)[/tex]: [tex]\(x + 10 = 9\)[/tex] and [tex]\(x + 5 = 4\)[/tex], both of which are positive and valid for logarithms.
Therefore, the valid solution is:
[tex]\[
x = -1.
\][/tex]
Hence, the value of [tex]\( x \)[/tex] that satisfies the equation [tex]\(\log_6(x + 10) + \log_6(x + 5) = 2\)[/tex] is:
[tex]\[
\boxed{-1}.
\][/tex]
