Question

解方程 log2(x^2-3)=2log4(6x-10)-1

写个过程 等于2

Answer 1

log2(x^2-3)=2log4(6x-10)-1
=》log2(x^2-3)=log4(6x-10)^2-1
=>log2(x^2-3)=log2(6x-10)-1
=》log2（x^2-3)=log2(6x-10)/2
=>x^2-3=(6x-10)/2
=>x^2-3x+2=0
=>x1=1(因为x^2-3<0,所以不符合要求），x2=2
=>x=2

Answer 2

log2(x^2-3)=2log4(6x-10)-1
log2(x^2-3)=log2(6x-10)-log2(2)=log2(3x-5)

x^2-3=3x-5>0,
x>5/3
x²-3x+2=0
(x-1)(x-2)=0
x1=1(不合舍去）
x2=2

Answer 3

log2(xˆ2-3)=2log4(6x-10)-1
log2(xˆ2-3)=2log2(6x-10)ˆ0.5-1
log2(xˆ2-3)=2*0.5log2(6x-10)-1
log2(xˆ2-3)=log2(6x-10)-1
log2(xˆ2-3)-log2(6x-10)=-1
log2[(xˆ2-3)/(6x-10)]=-1
(xˆ2-3)/(6x-10)=2ˆ-1
xˆ2-3=1/2*(6x-10)
接下来应该会做了吧