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

[tex] log_{x}(2(4x + 5)) \leqslant 1 \\log_{x}(2(4x + 5)) \leqslant log_{x}(x) \\ [/tex]

[tex]\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.



[/tex]

Ответ:
[tex]0 < x < 1 \\ [/tex]