2)log12(27)=a;log6(16)=?
log12(3^3)=3•log3(3)/log3(12)=3/(log3(3)+2log3(2))=3/(1+2log3(2))=a
3=a+2alog3(2)
log3(2)=(3-a)/2a
log6(16)=4log6(2)=4•log3(2)/log3(6)=
4log3(2)/(1+log3(2))=4(3-a)/2a:(1+(3-a)/2a)=
2(3-a)/a:(a+3)/2a=2(3-a)/a*2a/a+3=
4(3-a)/(3+a)
3)a)log30(x)=1/logx(30)=
1/(logx(5)+logx(2)+logx(3))
b)1/(log2(x)•log3(x)•log5(x))
*1/(logx(5)+logx(2)+logx(3))=
log2(x)•log3(x)+log3(x)•
log5(x)+log5(x)•log2(x)
c)(log2(x)•log3(x)+log3(x)•
log5(x)+log5(x)•log2(x)):
(log2(x)•log3(x)+log3(x)•
log5(x)+log5(x)•log2(x))
=1
B5 = b1·q^4 = -125·0,2^4= -125·(1/5)^4= -5^3 ·1/5^4= -1/5
(x2 - 6x - 21) (x3 - 6x) + 80 = 0
x4 - 12x3 +15x2 + 126x + 80 = 0
(x-8)(x+2)(x2 - 6x - 5) = 0
x-8 = 0
x+2=0
x2 - 6x - 5 =0
x=8
x= - 2
x2 - 6x = 5 x2 - 6x + 9 = 14 (x-3)2 = 14 х = 3 + √14 или х-3= - <span>√14</span>