Question

Срочно нужна помощь

Answer 1

$log_{ \frac{1}{2} } ^{2} 4x+ log_{2} \frac{x}{8} =7$

одз : x>0

$( log_{1/2} 4+ log_{1/2} x)^{2} + log_{2} x- log_{2} 8=7$

$(-2+ \frac{ log_{2} x}{ log_{2} \frac{1}{2} } )^{2} + log_{2} x-10=0$

$(2+ log_{2} x)^{2} + log_{2} x-10=0$

$(log_{2}x )^{2} +5* log_{2} x-6=0$

D=49

$1) log_{2} x=1 \\ x=2$


$2) log_{2} x=-6 \\ x= \frac{1}{64}$

ОТВЕТ  X₁=2   X₂=1/64