To solve the equation [tex]\(\log_6 z = 2 - \log_6 (z - 9)\)[/tex], we'll proceed step-by-step:
1. Given Equation:
[tex]\[
\log_6 z = 2 - \log_6(z - 9)
\][/tex]
2. Isolate the logarithms:
We'll try to combine the logarithmic terms.
[tex]\[
\log_6 z + \log_6 (z - 9) = 2
\][/tex]
3. Use the properties of logarithms:
We can combine the logarithms on the left side by using the property [tex]\(\log_b (a) + \log_b (c) = \log_b (a \cdot c)\)[/tex].
[tex]\[
\log_6 [z \cdot (z - 9)] = 2
\][/tex]
Simplifying inside the logarithm:
[tex]\[
\log_6 (z^2 - 9z) = 2
\][/tex]
4. Rewrite the equation using the definition of logarithms:
Recall that [tex]\(\log_b (a) = c\)[/tex] means [tex]\(b^c = a\)[/tex]. Here, [tex]\(b=6\)[/tex], [tex]\(c=2\)[/tex], and [tex]\(a=z^2 - 9z\)[/tex].
[tex]\[
6^2 = z^2 - 9z
\][/tex]
Simplifying the right-hand side:
[tex]\[
36 = z^2 - 9z
\][/tex]
5. Rearrange the equation:
Set the equation to equal zero.
[tex]\[
z^2 - 9z - 36 = 0
\][/tex]
6. Solve the quadratic equation:
Use the quadratic formula [tex]\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] where [tex]\(a = 1\)[/tex], [tex]\(b = -9\)[/tex], and [tex]\(c = -36\)[/tex].
[tex]\[
z = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-36)}}{2 \cdot 1}
\][/tex]
Simplify under the square root:
[tex]\[
z = \frac{9 \pm \sqrt{81 + 144}}{2}
\][/tex]
[tex]\[
z = \frac{9 \pm \sqrt{225}}{2}
\][/tex]
[tex]\[
z = \frac{9 \pm 15}{2}
\][/tex]
This gives us two potential solutions:
[tex]\[
z = \frac{9 + 15}{2} = \frac{24}{2} = 12
\][/tex]
[tex]\[
z = \frac{9 - 15}{2} = \frac{-6}{2} = -3
\][/tex]
7. Check the potential solutions:
We must check each potential solution in the original equation to ensure they are valid in the domain of the logarithm.
For [tex]\(z = 12\)[/tex]:
[tex]\[
\log_6 12 = 2 - \log_6 (12 - 9)
\][/tex]
[tex]\[
\log_6 12 = 2 - \log_6 3
\][/tex]
Verify [tex]\(\log_6 12\)[/tex] and [tex]\(2 - \log_6 3\)[/tex] can hold, as [tex]\(12\)[/tex] is in the domain of the logarithm.
For [tex]\(z = -3\)[/tex]:
[tex]\[
\log_6 (-3)
\][/tex]
Here, [tex]\(\log_6 (-3)\)[/tex] is undefined since the logarithm of a negative number does not exist in the real numbers.
8. Conclusion:
The only solution that is valid and within the domain is [tex]\(z = 12\)[/tex].
Therefore, the exact solution set is:
[tex]\[
\{12\}
\][/tex]
