Помогите решить уравнение, пожалуйста

Question 1

Question 2

ОДЗ:
$\begin{cases}x\ \textgreater \ 0;x\neq1\\1)log_x\sqrt{5x}\ \textgreater \ 0=\ \textgreater \ \left[\begin{gathered}\begin{cases}log_x\sqrt{5x}\ \textgreater \ 0\\x\ \textgreater \ 1\end{cases}\\\begin{cases}log_x\sqrt{5x}\ \textless \ 0\\0\ \textless \ x\ \textless \ 1\end{cases}\end{gathered}\\2)log_x5\ \textgreater \ 0=\ \textgreater \ \left[\begin{gathered}\begin{cases}log_x5\ \textgreater \ 0\\x\ \textgreater \ 1\end{cases}\\\begin{cases}log_x5\ \textless \ 0\\0\ \textless \ x\ \textless \ 1\end{cases}\end{gathered}\end{cases}$

$1)log_x\sqrt{5x}\ \textgreater \ 0(x\ \textgreater \ 1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ log_x\sqrt{5x}\ \textless \ 0(0\ \textless \ x\ \textless \ 1)\\\sqrt{5x}\ \textgreater \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{5x}\ \textgreater \ 1\\\sqrt{5x}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5x=1\\5x=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5x=1\\x=0,2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0,2$
(1)/////////////>x
(0,2)///////////////////////////////>x

(0)///////////////////////(1) >x
//////////////(0,2)
$x\in(0;0,2)\cup(1;+\infty)$

$log_x5\ \textgreater \ 0(x\ \textgreater \ 1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ log_x5\ \textless \ 0(0\ \textless \ x\ \textless \ 1)\\5\ \textgreater \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5\ \textless \ 1\\x\in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in\varnothing$
(1)/////////////>x
////////////////////////////////////////////////////>x

(0)///////////////////////(1) >x
>x
$x\in(1;+\infty)$

В общем получаем ОДЗ: x>1

$\sqrt{log_x\sqrt{5x}}=log_x5\\log_x\sqrt{5x}=log_x^25\\\frac{1}{2}log_x5x=log_x^25|*2\\log_x5x=2log_x^25\\log_x5+log_xx=2log_x^25\\2log_x^25-log_x5-1=0\\log_x5_{1,2}=\frac{1^+_-3}{4}\\log_x5=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ log_x5=-\frac{1}{2}\\x=5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\sqrt x}=5\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt x=\frac{1}{5}\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\frac{1}{25}$
$.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in\varnothing$
Ответ: х=5

Alexаndr_zn · Answer 1 · 2018-05-15T10:23:26+0000

ОДЗ:
$\begin{cases}x\ \textgreater \ 0;x\neq1\\1)log_x\sqrt{5x}\ \textgreater \ 0=\ \textgreater \ \left[\begin{gathered}\begin{cases}log_x\sqrt{5x}\ \textgreater \ 0\\x\ \textgreater \ 1\end{cases}\\\begin{cases}log_x\sqrt{5x}\ \textless \ 0\\0\ \textless \ x\ \textless \ 1\end{cases}\end{gathered}\\2)log_x5\ \textgreater \ 0=\ \textgreater \ \left[\begin{gathered}\begin{cases}log_x5\ \textgreater \ 0\\x\ \textgreater \ 1\end{cases}\\\begin{cases}log_x5\ \textless \ 0\\0\ \textless \ x\ \textless \ 1\end{cases}\end{gathered}\end{cases}$

$1)log_x\sqrt{5x}\ \textgreater \ 0(x\ \textgreater \ 1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ log_x\sqrt{5x}\ \textless \ 0(0\ \textless \ x\ \textless \ 1)\\\sqrt{5x}\ \textgreater \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{5x}\ \textgreater \ 1\\\sqrt{5x}=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5x=1\\5x=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5x=1\\x=0,2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0,2$
(1)/////////////>x
(0,2)///////////////////////////////>x

(0)///////////////////////(1) >x
//////////////(0,2)
$x\in(0;0,2)\cup(1;+\infty)$

$log_x5\ \textgreater \ 0(x\ \textgreater \ 1)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ log_x5\ \textless \ 0(0\ \textless \ x\ \textless \ 1)\\5\ \textgreater \ 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5\ \textless \ 1\\x\in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in\varnothing$
(1)/////////////>x
////////////////////////////////////////////////////>x

(0)///////////////////////(1) >x
>x
$x\in(1;+\infty)$

В общем получаем ОДЗ: x>1

$\sqrt{log_x\sqrt{5x}}=log_x5\\log_x\sqrt{5x}=log_x^25\\\frac{1}{2}log_x5x=log_x^25|*2\\log_x5x=2log_x^25\\log_x5+log_xx=2log_x^25\\2log_x^25-log_x5-1=0\\log_x5_{1,2}=\frac{1^+_-3}{4}\\log_x5=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ log_x5=-\frac{1}{2}\\x=5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{\sqrt x}=5\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt x=\frac{1}{5}\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=\frac{1}{25}$
$.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x\in\varnothing$
Ответ: х=5