Question

log2(3x-1)-log2(4-x)=4-log2(x-1)

Answer 1

log2(3x-1)-log2(4-x)=4-log2(x-1)
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ОДЗ:
{3x-1>0; 3x>1; x>1/3
{4-x>0; -x>-4; x<4
{x-1>0; x>1
Решение ОДЗ: x e (1;4)
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log2(3x-1)-log2(4-x)=log2(16)-log2(x-1)
log2[(3x-1)/(4-x)] = log2[16/(x-1)]
(3x-1)/(4-x)= 16/(x-1)
16(4-x)=(3x-1)(x-1)
64-16x=3x^2-4x+1
64-16x-3x^2+4x-1=0
-3x^2-12x+63=0
3x^2+12x-63=0 |:3
x^2+4x-21=0
D=4^2-4*1*(-21)=100
x1=(-4-10)/2=-7 - не подходит по ОДЗ
x2=(-4+10)/2=3
Ответ:3