Пошаговое объяснение:
[tex] log_{5}(3 - 8x) > 0 \\ 3 - 8x > {5}^{0} \\ 3 - 8x > 1 \\ 2 - 8x > 0 \\ x > \frac{1}{4} [/tex]
[tex] log_{3}(3 x - 1) < log_{3}(2x + 3) \\ 3x - 1 < 2x + 3 \\ x < 4[/tex]
[tex] log_{0.5}(6 - x) > log_{0.5}(4) \\ 6 - x > 4 \\ x < 2[/tex]
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Answers & Comments
Пошаговое объяснение:
1)
[tex] log_{5}(3 - 8x) > 0 \\ 3 - 8x > {5}^{0} \\ 3 - 8x > 1 \\ 2 - 8x > 0 \\ x > \frac{1}{4} [/tex]
2)
[tex] log_{3}(3 x - 1) < log_{3}(2x + 3) \\ 3x - 1 < 2x + 3 \\ x < 4[/tex]
3)
[tex] log_{0.5}(6 - x) > log_{0.5}(4) \\ 6 - x > 4 \\ x < 2[/tex]