Let's start with the equation [tex]\(\log_6 x + \log_6 (x+5) = 2\)[/tex].
### Step-by-Step Solution:
1. Combine the logarithmic terms:
[tex]\[ \log_6 x + \log_6 (x+5) = \log_6 [x \cdot (x+5)] \][/tex]
This uses the logarithmic property: [tex]\(\log_b a + \log_b c = \log_b (a \cdot c)\)[/tex].
2. Simplify the equation:
[tex]\[ \log_6 [x \cdot (x+5)] = 2 \][/tex]
3. Convert the logarithmic equation to an exponential equation:
If [tex]\(\log_b a = c\)[/tex], then [tex]\(a = b^c\)[/tex]. Applying this property:
[tex]\[ x \cdot (x+5) = 6^2 \][/tex]
4. Simplify further:
[tex]\[ x(x + 5) = 36 \][/tex]
[tex]\[ x^2 + 5x - 36 = 0 \][/tex]
5. Solve the quadratic equation:
To solve the quadratic equation [tex]\(x^2 + 5x - 36 = 0\)[/tex], we can use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = -36\)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-36) \][/tex]
[tex]\[ \Delta = 25 + 144 \][/tex]
[tex]\[ \Delta = 169 \][/tex]
Find the roots:
[tex]\[ x = \frac{-5 \pm \sqrt{169}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-5 \pm 13}{2} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{-5 + 13}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ x_2 = \frac{-5 - 13}{2} = \frac{-18}{2} = -9 \][/tex]
6. Check the solutions:
We must validate that they are within the domain of the original logarithmic function. Logarithms are defined for positive real numbers.
- For [tex]\(x = 4\)[/tex]:
Both [tex]\( \log_6 x = \log_6 4 \)[/tex] and [tex]\( \log_6 (x+5) = \log_6 9 \)[/tex] are defined.
- For [tex]\(x = -9\)[/tex]:
[tex]\(\log_6 (-9)\)[/tex] is not defined because the logarithm of a negative number is not a real number.
7. Identify valid solutions:
From the initial options [tex]\(-12, -9, 4, 32, 36\)[/tex], we see that the valid solution [tex]\(x = 4\)[/tex] is within these options.
### Conclusion:
The potential solution for the equation [tex]\(\log_6 x + \log_6 (x+5) = 2\)[/tex] from the given options is:
[tex]\[ x = 4 \][/tex]
