Answer:
[tex]\sf x = 12[/tex]
Step-by-step explanation:
To solve the equation [tex]\sf \log_6(x) + \log_6(x-9) = 2[/tex], we can apply the logarithmic rule for the sum of logarithms:
[tex]\sf \log_b(M) + \log_b(N) = \log_b(M \times N)[/tex]
Using this rule, we can rewrite the equation as:
[tex]\sf \log_6(x \times (x-9)) = 2[/tex]
Now, we simplify inside the logarithm:
[tex]\sf x(x-9) = 6^2[/tex]
[tex]\sf x^2 - 9x = 36[/tex]
Rearranging the equation into a quadratic form:
[tex]\sf x^2 - 9x - 36 = 0[/tex]
Now, we can factor or use the quadratic formula to solve for [tex]\sf x[/tex]. In this case, let's use the quadratic formula:
[tex]\sf x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
Where [tex]\sf a = 1[/tex], [tex]\sf b = -9[/tex], and [tex]\sf c = -36[/tex].
[tex]\sf x = \dfrac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(-36)}}{2(1)}[/tex]
[tex]\sf x = \dfrac{9 \pm \sqrt{81 + 144}}{2}[/tex]
[tex]\sf x = \dfrac{9 \pm \sqrt{225}}{2}[/tex]
[tex]\sf x = \dfrac{9 \pm 15}{2}[/tex]
So, the solutions for [tex]\sf x[/tex] are:
[tex]\sf x_1 = \dfrac{9 + 15}{2} = 12[/tex]
[tex]\sf x_2 = \dfrac{9 - 15}{2} = -3[/tex]
However, we need to verify whether both solutions are valid. In the context of logarithms, the argument of a logarithm cannot be negative.
Therefore, [tex]\sf x = -3[/tex] is an extraneous solution.
Thus, the only valid solution is [tex]\sf x = 12[/tex].
