Найти корень уравнения log3(4x+1)=2-log3(x-1)

Question 1

Question 2

$log_{3}(4x+1)=2-log_{3}(x-1)$

ОДЗ:

0} \atop {x-1>0}} \right. " alt=" \left \{ {{4x+1>0} \atop {x-1>0}} \right. " align="absmiddle" class="latex-formula">

- \frac{1}{4} } \atop {x>1}} \right. " alt=" \left \{ {{x>- \frac{1}{4} } \atop {x>1}} \right. " align="absmiddle" class="latex-formula">

1" alt="x>1" align="absmiddle" class="latex-formula">

$log_{3}(4x+1)=log_{3}9-log_{3}(x-1)$
$log_{3}(4x+1)=log_{3}( \frac{9}{x-1})$
$4x+1=\frac{9}{x-1}$
$(4x+1)(x-1)=9$
$4x^{2}-4x+x-1-9=0$
$4x^{2}-3x-10=0, D=9+4*4*10=169=13^{2}$
$x_{1}= \frac{3-13}{8}=-\frac{10}{8}=-\frac{5}{4}=-1.25<1$ - посторонний корень

1" alt="x_{2}= \frac{3+13}{8}=\frac{16}{8}=2>1" align="absmiddle" class="latex-formula"> - ответ

kalbim_zn · Answer 1 · 2018-03-20T04:24:45+0000

$log_{3}(4x+1)=2-log_{3}(x-1)$

ОДЗ:

0} \atop {x-1>0}} \right. " alt=" \left \{ {{4x+1>0} \atop {x-1>0}} \right. " align="absmiddle" class="latex-formula">

- \frac{1}{4} } \atop {x>1}} \right. " alt=" \left \{ {{x>- \frac{1}{4} } \atop {x>1}} \right. " align="absmiddle" class="latex-formula">

1" alt="x>1" align="absmiddle" class="latex-formula">

$log_{3}(4x+1)=log_{3}9-log_{3}(x-1)$
$log_{3}(4x+1)=log_{3}( \frac{9}{x-1})$
$4x+1=\frac{9}{x-1}$
$(4x+1)(x-1)=9$
$4x^{2}-4x+x-1-9=0$
$4x^{2}-3x-10=0, D=9+4*4*10=169=13^{2}$
$x_{1}= \frac{3-13}{8}=-\frac{10}{8}=-\frac{5}{4}=-1.25<1$ - посторонний корень

1" alt="x_{2}= \frac{3+13}{8}=\frac{16}{8}=2>1" align="absmiddle" class="latex-formula"> - ответ