log₈₁(x+6)+log_((x²+12x+36))3 ≥ 3/4

Answer

Explanation:

Step1: Simplify the second logarithm's base

Note that (x^{2}+12x + 36=(x + 6)^{2}), so the inequality becomes (\log_{81}(x + 6)+\log_{(x + 6)^{2}}3\geq\frac{3}{4}). Using the change - of - base formula (\log_{a^{n}}b=\frac{1}{n}\log_{a}b), we have (\log_{(x + 6)^{2}}3=\frac{1}{2}\log_{x + 6}3=\frac{1}{2}\cdot\frac{\log_{3}3}{\log_{3}(x + 6)}=\frac{1}{2\log_{3}(x + 6)}) (by the change - of - base formula (\log_{a}b=\frac{\log_{c}b}{\log_{c}a})). Also, (\log_{81}(x + 6)=\frac{\log_{3}(x + 6)}{\log_{3}81}=\frac{\log_{3}(x + 6)}{4}) (since (81 = 3^{4})). Let (t=\log_{3}(x + 6)), then the inequality is (\frac{t}{4}+\frac{1}{2t}\geq\frac{3}{4}), where (x+6>0) (so (x>-6)) and (x + 6\neq1) (so (x\neq - 5)) (because the base of a logarithm must be positive and not equal to 1).

Step2: Solve the rational inequality

Multiply both sides of (\frac{t}{4}+\frac{1}{2t}-\frac{3}{4}\geq0) by (4t) (note that we need to consider the sign of (t) because when multiplying an inequality by a variable, the direction of the inequality may change). Case 1: (t>0) (t^{2}+2 - 3t\geq0), that is (t^{2}-3t + 2\geq0). Factor the quadratic: ((t - 1)(t - 2)\geq0). The solutions of the quadratic equation ((t - 1)(t - 2)=0) are (t = 1) and (t = 2). For the quadratic function (y=(t - 1)(t - 2)=t^{2}-3t + 2), the parabola opens upwards. So the inequality ((t - 1)(t - 2)\geq0) holds when (t\leq1) or (t\geq2). Combining with (t>0), we have (0<t\leq1) or (t\geq2). Case 2: (t<0) (t^{2}+2-3t\leq0) (because we multiply by a negative number (4t), the inequality sign flips), (t^{2}-3t + 2\leq0), ((t - 1)(t - 2)\leq0). The solution of ((t - 1)(t - 2)\leq0) is (1\leq t\leq2). But we assumed (t<0), so there is no solution in this case.

Step3: Substitute back (t = \log_{3}(x + 6))

When (0<\log_{3}(x + 6)\leq1): Using the property of logarithms, if (y = \log_{a}x) ((a>1)), then (x) is increasing with (y). So (3^{0}<x + 6\leq3^{1}), that is (1<x + 6\leq3), so (-5<x\leq - 3). But we need to exclude (x=-5) (from the base of the second logarithm), so (-5<x\leq - 3).
When (\log_{3}(x + 6)\geq2): (x + 6\geq3^{2}=9), so (x\geq3).

Answer:

The solution of the inequality (\log_{81}(x + 6)+\log_{x^{2}+12x + 36}3\geq\frac{3}{4}) is ((-5,-3]\cup[3,+\infty))