Решите уравнение: [tex] log_{2} x^{2} (3x+1)-3 log_{1/2} \frac{4}{3x+1} = \frac{2}{7} ^{log_{2/7}1,5+log_{2/7} 4 } [/tex]. Created by emilynight1. matematika-ru

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August 2022 1 11 Report

Решите уравнение:
[tex] log_{2} x^{2} (3x+1)-3 log_{1/2} \frac{4}{3x+1} = \frac{2}{7} ^{log_{2/7}1,5+log_{2/7} 4 } [/tex]

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hote

Verified answer

$log_2(x^2*(3x+1))-3*log_{1/2}( \frac{4}{3x+1})= \frac{2}{7}^{log_{2*7}1.5+log_{2/7}4$

$ODZ:
 \left \{ {{(3x+1)\ \textgreater \ 0} \atop {3x+1 \neq 0}} \right.

 \left \{ {{x\ \textgreater \ -1/3} \atop {x \neq -1/3}} \right.

(-1/3: +oo)$

$log_2x^2+log_2(3x+1)-3Log_{1/2^{-1}} \frac{4}{3x+1}= \frac{2}{7}^{log_{2/7}1.5*4$

$log_2 x^2+log_2(3x+1)+3(log_2 \frac{4}{3x+1})= \frac{2}{7}^{log_{2/7}6$

$log_2x^2+log_2(3x+1)+3log_24-3log_2(3x+1)=6$

$log_2x^2+log_2(3x+1)+6-3log_2(3x+1)=6$

$log_2x^2-2log(3x+1)=0$

$log_2x^2-log_2(3x+1)^2=0$

$log_2( \frac{x^2}{(3x+1)^2})=0$

$\frac{x^2}{(3x+1)^2}=2^0$

$\frac{x^2}{(3x+1)^2}=1$

$x^2=(3x+1)^2

x^2=9x^2+6x+1

8x^2+6x+1=0

D=36-32=4=2^2

x= (-6-2)/16=-8/16=-1/2

x= (-6+2)/16=-4/16=-1/4$

$x= -1/2 \neq (-1/3; +oo)$
не подходит

Ответ
$x= - \frac{1}{4}$

1 votes Thanks 1

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emilynight1 August 2022 | 0 Ответы

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