作业帮对数解方程 log2(x-1)-log4(x+2)+1=0 2和四是底 括号里的是真数
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对数解方程 log2(x-1)-log4(x+2)+1=0 2和四是底 括号里的是真数
对数解方程 log2(x-1)-log4(x+2)+1=0 2和四是底 括号里的是真数
对数解方程 log2(x-1)-log4(x+2)+1=0 2和四是底 括号里的是真数
log2(x-1)-log4(x+2)+1=0
(x-1)^2/(x+2)=1/4
整理,得X^2-2.25x+1.5=0
得,X=0.25或2
检验,得X=0.25不符合题意,X=2符合题意
所以,X=2
lg(x-1)-lg(x+2)+lg2=0
2x-2=x+2
x=4
log2(x-1)-log4(x+2)+1=0
∵log2(x-1)-1/2*log2(x+2)+1=0
∴log2[(x-1)/√（x+2)]=-1
∴[(x-1)/√（x+2)]=1/2
∴4(x-1)^2=x+2
∴x=1/4或x=2
又∵ x-1＞0，x+2＞0
∴x＞1
∴ x=2
log2(x-1)-log4(x+2)+1=0
log2(x-1)-log2^2(x+2)+1=0
log2(x-1)-1/2log2(x+2)+1=0
2log2(x-1)-log2(x+2)+2=0
log2(x-1)^2-log2(x+2)=-2
log2[(x-1)^2/(x+2)]=log2[(2^(-2)]
(x-1)^2/(x+2)=2...
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log2(x-1)-log4(x+2)+1=0
log2(x-1)-log2^2(x+2)+1=0
log2(x-1)-1/2log2(x+2)+1=0
2log2(x-1)-log2(x+2)+2=0
log2(x-1)^2-log2(x+2)=-2
log2[(x-1)^2/(x+2)]=log2[(2^(-2)]
(x-1)^2/(x+2)=2^(-2)=1/4
4(x-1)^2=x+2
4(x^2-2x+1)=x+2
4x^2-8x+4-x-2=0
4x^2-9x+2=0
(4x-1)(x-2)=0
x=1/4或x=2
但log2(x-1)中x-1>0
x>1
所以x=2
收起
log2(x-1)-log4(x+2)+1=0
log2(x-1)-(1/2)log2(x+2)+1=0
2log2(x-1)-log2(x+2)+2=0
log2(x-1)^2-log2(x+2)+2=0
(x-1)^2/(x+2)=1/4
4(x-1)^2=x+2
x1=1/4(不合题意，舍去）
x2=2
log2(x-1)-log4(x+2)+1
=log2(x-1)-(1/2)log2(x+2)+1
=log2[(x-1)/(x+2)^(1/2)]+1
=0
所以 log2[(x-1)/(x+2)^(1/2)]=-1=log2(1/2)
即 (x-1)/(x+2)^(1/2)=1/2
2(x-1)=(x+2)^(1/2)
4(x-1)^2...
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log2(x-1)-log4(x+2)+1
=log2(x-1)-(1/2)log2(x+2)+1
=log2[(x-1)/(x+2)^(1/2)]+1
=0
所以 log2[(x-1)/(x+2)^(1/2)]=-1=log2(1/2)
即 (x-1)/(x+2)^(1/2)=1/2
2(x-1)=(x+2)^(1/2)
4(x-1)^2=x+2
4x^2-9+2=0
解得x=1/4 或 x=2
因为方程中括号里是真数，x-1>=0，
所以x=1/4舍去
所以x=2
收起