Question

Log0.3(−x+2)≤log0.3(2x−2)

Answer 1

Ответ:

$(1;1\frac{1}{3}]$

Пошаговое объяснение:

$log_{0,3}(-x+2)\leq log_{0,3}(2x-2)\\\\OD3:\\\left \{ {{-x+2>0} \atop {2x-2>0}} \right.=>\left \{ {{-x>-2} \atop {2x>2}} \right.=>\left \{ {{x1}} \right.=>x\in(1;2)\\\\0$

Answer 2

To solve the inequality log₀.₃(-x + 2) ≤ log₀.₃(2x - 2), you can use the properties of logarithms. In this case, since both sides have the same base (0.3), you can eliminate the logarithms by equating the arguments (what's inside the logarithms) and then solving for x.

1. Start by setting the arguments equal to each other:

   -x + 2 = 2x - 2

2. Now, solve for x. First, add x to both sides:

   2x - x + 2 = -2

3. Simplify:

   x + 2 = -2

4. Subtract 2 from both sides:

   x = -2 - 2

5. Continue simplifying:

   x = -4

So, the solution to the inequality log₀.₃(-x + 2) ≤ log₀.₃(2x - 2) is x ≤ -4.