Question

Log1/6(10 – x) + log1/6(x – 3) ≥ - 1.

Answer 1

Log₁/₆(10-x)+log₁/₆(x-3)≥-1
ОДЗ:  10-x>0  x<10   x-3>0  x>3   x∈(3;10)
log₁/₆(10-x)(x-3)≥log₁/₆6
-x²+13x-30≤6  I×(-1)
x²-13x+36≥0  D=25
x₁=4    x₂=9
(x-4)(x-9)≥0
-∞_____+_____4_____-_____9_____+______+∞
x∈(-∞;4]U[9;+∞)
Учитывая ОДЗ х∈(3;4]U[9;10).

Answer 2

To solve the logarithmic inequality log₁/₆(10 - x) + log₁/₆(x - 3) ≥ -1, we need to apply some properties of logarithms.

First, let's combine the two logarithms into a single logarithm using the product rule:

log₁/₆[(10 - x)(x - 3)] ≥ -1

Next, let's convert the inequality into exponential form:

₁/₆[(10 - x)(x - 3)] ≥ ₁/₆

Now, we can cancel out the base 1/6 on both sides by raising it to the same power:

(10 - x)(x - 3) ≥ 1

Expand the equation:

10x - 3x - 10(-3) + 3(10) ≥ 1

7x + 30 ≥ 1

Subtract 30 from both sides:

7x ≥ -29

Divide both sides by 7:

x ≥ -29/7

Therefore, the solution to the inequality log₁/₆(10 - x) + log₁/₆(x - 3) ≥ -1 is x ≥ -29/7.