Решить неравенства и указать все его целые решения:1) log3(x)>log3(5-x)2)...

Question 1

Решить неравенства и указать все его целые решения:
1) log3(x)>log3(5-x)
2) log1/7(2x+3)<log1/7(3x-2)

Question 2

1)
$log_3x\ \textgreater \ log_3(5-x)$
ОДЗ:
$\left \{ {{x\ \textgreater \ 0} \atop {5-x\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ 0} \atop {x\ \textless \ 5}} \right.$
$x$ ∈ $(0;5)$

$x\ \textgreater \ 5-x$
$x+x\ \textgreater \ 5$
$2x\ \textgreater \ 5$
$x\ \textgreater \ 2.5$

------------(2.5)-------------------
//////////////////////
----(0)--------------(5)-----------
///////////////
$x$ ∈ $(2.5;5)$

Ответ: целые решения: 3; 4

2)
$log_ \frac{1}{7} (2x+3)\ \textless \ log_ \frac{1}{7} (3x-2)$
JLP^
$\left \{ {{2x+3\ \textgreater \ 0} \atop {3x-2\ \textgreater \ 0}} \right.$
$\left \{ {{2x\ \textgreater \ -3} \atop {3x\ \textgreater \ 2}} \right.$
$\left \{ {{x\ \textgreater \ -1.5} \atop {x\ \textgreater \ \frac{2}{3} }} \right.$
$x$ ∈ $( \frac{2}{3} ;+$ ∞ $)$

$2x+3\ \textgreater \ \ 3x-2$
$2x-3x\ \textgreater \ \ -2-3$
$-x\ \textgreater \ -5$
$x\ \textless \ 5$

----------(2/3)------------------
//////////////////////
-----------------------(5)-------
//////////////////////////
$x$ ∈ $( \frac{2}{3} ;5)$

Ответ: целые решения: 1; 2; 3; 4

Luluput_zn · Answer 1 · 2018-05-09T22:03:05+0000

1)
$log_3x\ \textgreater \ log_3(5-x)$
ОДЗ:
$\left \{ {{x\ \textgreater \ 0} \atop {5-x\ \textgreater \ 0}} \right.$
$\left \{ {{x\ \textgreater \ 0} \atop {x\ \textless \ 5}} \right.$
$x$ ∈ $(0;5)$

$x\ \textgreater \ 5-x$
$x+x\ \textgreater \ 5$
$2x\ \textgreater \ 5$
$x\ \textgreater \ 2.5$

------------(2.5)-------------------
//////////////////////
----(0)--------------(5)-----------
///////////////
$x$ ∈ $(2.5;5)$

Ответ: целые решения: 3; 4

2)
$log_ \frac{1}{7} (2x+3)\ \textless \ log_ \frac{1}{7} (3x-2)$
JLP^
$\left \{ {{2x+3\ \textgreater \ 0} \atop {3x-2\ \textgreater \ 0}} \right.$
$\left \{ {{2x\ \textgreater \ -3} \atop {3x\ \textgreater \ 2}} \right.$
$\left \{ {{x\ \textgreater \ -1.5} \atop {x\ \textgreater \ \frac{2}{3} }} \right.$
$x$ ∈ $( \frac{2}{3} ;+$ ∞ $)$

$2x+3\ \textgreater \ \ 3x-2$
$2x-3x\ \textgreater \ \ -2-3$
$-x\ \textgreater \ -5$
$x\ \textless \ 5$

----------(2/3)------------------
//////////////////////
-----------------------(5)-------
//////////////////////////
$x$ ∈ $( \frac{2}{3} ;5)$

Ответ: целые решения: 1; 2; 3; 4