1) log2 28+2 log2 12- log2 63 2)решить неравенство: (log2x)^2-3 log2 x≤4 3) log3(2x+5)=4

1) log₂28+2*log₂12-log₂63=log₂28+log₂12²-log₂63=log₂(28*144/63)=
=log₂(4032/63)=log₂64=log₂2⁶=6*log₂2=6.
2)
(log₂x)²-3*log₂x≤4    ОДЗ: x>0
Пусть log₂x=t  ⇒
t²-3t≤4
t²-3t-4≤0
t²-3t-4=0   D=25
t₁=-1        t₂=4   ⇒
 (t+1)(t-4)≤0
 (log₂x+1)(log₂x-4)≤0     
log₂x-4=0    log₂x=4    x₁=2⁴=16    
log₂x+1=0   log₂x=-1   x₂=2⁻¹=1/2   ⇒
(x-1/2)(x-16)≤0
-∞+1/2-16+∞
x∈[1/2;16].
3)
log₃(2x+5)=4
2x+5=3⁴
2x+5=81
2x=76
x=38.