Question

Log3(x^2+2)- log 3 (x^2-x+12) >= log3(1+ 1/x)

Answer 1

$log_3(x^2+2)- log_3 (x^2-x+12) \geq log_3\Big(1+ \dfrac{1}{x}\Big)$

ОДЗ : 1) x²+2>0    ⇒   x∈R

         2) x²-x+12>0   ⇒  D=1-4·12=-47<0   ⇒  x∈R

    3) $1+\dfrac{1}{x}>0~~\Leftrightarrow~~\dfrac{x+1}{x}>0~~\Rightarrow~~x\in (-\infty;-1)\cup(0;+\infty)$

ОДЗ :  x ∈ (-∞; -1) ∪ (0; +∞)

$log_3(x^2+2)- log_3 (x^2-x+12) \geq log_3\Big(\dfrac{x+1}{x}\Big)\\ \\ log_3\Big(\dfrac{x^2+2}{x^2-x+12}\Big) \geq log_3\Big(\dfrac{x+1}{x}\Big)\\ \\ 3>1~~~\Rightarrow\\ \\ \dfrac{x^2+2}{x^2-x+12} \geq \dfrac{x+1}{x}\\ \\ \\ \dfrac{x^2+2}{x^2-x+12} - \dfrac{x+1}{x}\geq 0\\ \\ \\ \dfrac{x(x^2+2)-(x+1)((x^2-x+12))}{(x^2-x+12)x} \geq 0\\ \\ \\ \dfrac{x^3+2x-(x^3+x^2-x^2-x+12x+12)}{(x^2-x+12)x} \geq 0\\ \\ \\ \dfrac{x^3+2x-x^3-11x-12}{(x^2-x+12)x} \geq 0$

$></p> <p>x² - x + 12 > 0   всегда, так как D < 0    ⇒</p> <p><img src=$

Метод интервалов : x₁ = $-1\frac{1}{3}$;   x₂ = 0

+++++++++++ $[-1\frac{1}{3}]$ ----------- (0) +++++++++++> x

$\boldsymbol{x \in [-1\frac{1}{3};0)}$

С учётом ОДЗ :    $\boxed{\boldsymbol{x \in [-1\frac{1}{3};-1)}}$

Answer 2

To solve the inequality log₃(x²+2) - log₃(x²-x+12) ≥ log₃(1 + 1/x), we need to use logarithmic properties and algebraic manipulation. First, let's rewrite the inequality using the properties of logarithms:

log₃(x²+2) - log₃(x²-x+12) ≥ log₃(1 + 1/x)

Step 1: Combine the logarithms on the left side using the quotient rule of logarithms:

log₃[(x²+2)/(x²-x+12)] ≥ log₃(1 + 1/x)

Step 2: Remove the logarithms using the exponential function:

(x²+2)/(x²-x+12) ≥ 1 + 1/x

Step 3: Eliminate the fraction by multiplying both sides by x(x²-x+12):

x(x²+2) ≥ x(x²-x+12) + (x²-x+12)

Step 4: Expand and simplify:

x³ + 2x ≥ x³ - x² + 12x + x² - x + 12

Step 5: Cancel out x³ terms on both sides:

2x ≥ 11x + 12

Step 6: Move all x terms to the left side and the constants to the right side:

2x - 11x ≥ 12

Step 7: Combine like terms:

-9x ≥ 12

Step 8: Divide both sides by -9, but since we are dividing by a negative number, the inequality sign flips:

x ≤ 12/-9

Step 9: Simplify the fraction:

x ≤ -4/3

So, the solution to the inequality is x ≤ -4/3.