Log(x-основание) 2(4x+5) меньше или равно 1

Question

Log(x-основание) 2(4x+5) меньше или равно 1

Answer 1

Решение во вложение . Приложение photomath

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Answer 2

1} \atop {8x + 10 \leqslant x}} \right. \\ \\ \left \{ {{0 < x < 1} \atop { 7x } \geqslant \: - 10} \right. and \\ \left \{ {{x > 1} \atop {7x \leqslant - 10}} \right. \\ \\ \\\left \{ {{0 < x < 1} \atop { x } \geqslant \: - 1\frac{3}{7} } \right. and \\ \left \{ {{x > 1} \atop {x \leqslant - 1 \frac{3}{7} }} \right. \\ \\ \\ \left \{ {{0 < x < 1} \atop { x } \geqslant \: - 1\frac{3}{7} } \right.\\\\" alt="\left \{ {{0 < x < 1} \atop { 8x + 10 } \geqslant \: x } \right. and \\ \left \{ {{x > 1} \atop {8x + 10 \leqslant x}} \right. \\ \\ \left \{ {{0 < x < 1} \atop { 7x } \geqslant \: - 10} \right. and \\ \left \{ {{x > 1} \atop {7x \leqslant - 10}} \right. \\ \\ \\\left \{ {{0 < x < 1} \atop { x } \geqslant \: - 1\frac{3}{7} } \right. and \\ \left \{ {{x > 1} \atop {x \leqslant - 1 \frac{3}{7} }} \right. \\ \\ \\ \left \{ {{0 < x < 1} \atop { x } \geqslant \: - 1\frac{3}{7} } \right.\\\\" align="absmiddle" class="latex-formula">

Ответ:

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