(3*㏒_2(x))/(㏒_2(x)-1)=(㏒_2(x)-2)/(㏒_2(x)). Created by elvira1234123. algebra-ru

ОДЗ: {x>0 {x>0 {x>0{log₂x-1≠0 => {log₂x≠1 => {x≠2 => x∈(0;1)U(1;2)U(2;+∞){log₂x≠0 {x≠2⁰ {x≠1Ответ: √2; 1/4

(㏒_2(x))...

elvira1234123 @elvira1234123

August 2022 1 10 Report

(3*㏒_2(x))/(㏒_2(x)-1)=(㏒_2(x)-2)/(㏒_2(x))

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Verified answer

$\frac{3log_2x}{log_2x-1} = \frac{log_2x-2}{log_2x}$

ОДЗ:
{x>0 {x>0 {x>0
{log₂x-1≠0 => {log₂x≠1 => {x≠2 => x∈(0;1)U(1;2)U(2;+∞)
{log₂x≠0 {x≠2⁰ {x≠1

$3log^2_2x=log_2^2x-log_2x-2log_2x+2\\2log^2_2x+3log_2x-2=0\\y=log_2x\\2y^2+3y-2=0\\D=3^2-4*2*(-2)=9+16=25=5^2\\y_1=(-3+5)/(2*2)=2/4=1/2\\y_2=(-3-5)/(2*2)=-8/4=-2\\\\log_2x=1/2\\x=2^{1/2}\\x_1= \sqrt{2} \; \; (\in OD3)\\\\log_2x=-2\\x=2^{-2}\\x_2=1/4\; \; (\in OD3)$

Ответ: √2; 1/4

2 votes Thanks 1

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