To solve the equation [tex]\(\log_5(x + 2) = 1 - \log_5(x + 6)\)[/tex], we'll proceed with the following steps:
1. Isolate the logarithms on one side:
[tex]\[
\log_5(x + 2) + \log_5(x + 6) = 1
\][/tex]
2. Use the properties of logarithms:
Recall the property of logarithms that states [tex]\(\log_b(a) + \log_b(c) = \log_b(ac)\)[/tex]. Applying this property, we have:
[tex]\[
\log_5((x + 2)(x + 6)) = 1
\][/tex]
3. Convert the logarithmic equation to an exponential equation:
Recall that if [tex]\(\log_b(y) = c\)[/tex], then [tex]\(b^c = y\)[/tex]. So, we get:
[tex]\[
5^1 = (x + 2)(x + 6)
\][/tex]
4. Simplify and solve the quadratic equation:
[tex]\[
5 = (x + 2)(x + 6)
\][/tex]
Expand the right-hand side:
[tex]\[
5 = x^2 + 6x + 2x + 12
\][/tex]
[tex]\[
5 = x^2 + 8x + 12
\][/tex]
Subtract 5 from both sides to set the equation to zero:
[tex]\[
0 = x^2 + 8x + 7
\][/tex]
5. Factor the quadratic equation:
We need to factor [tex]\(x^2 + 8x + 7\)[/tex]. Looking for two numbers that multiply to 7 and add up to 8, we get 1 and 7.
[tex]\[
0 = (x + 1)(x + 7)
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
[tex]\[
x + 1 = 0 \quad \text{or} \quad x + 7 = 0
\][/tex]
[tex]\[
x = -1 \quad \text{or} \quad x = -7
\][/tex]
7. Check for valid solutions:
* For [tex]\(x = -1\)[/tex]:
[tex]\[
\log_5(-1 + 2) = \log_5(1) = 0
\][/tex]
[tex]\[
1 - \log_5(-1 + 6) = 1 - \log_5(5) = 1 - 1 = 0
\][/tex]
Both sides are equal, so [tex]\(x = -1\)[/tex] is a valid solution.
* For [tex]\(x = -7\)[/tex]:
[tex]\[
\log_5(-7 + 2) = \log_5(-5)
\][/tex]
The logarithm of a negative number is not defined, so [tex]\(x = -7\)[/tex] is not a valid solution.
Thus, the only valid solution is:
[tex]\[
x = -1
\][/tex]
