y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x
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y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x|] /2 这几步不太理解,
注:题中是以4为底和以2为底
注:题中是以4为底和以2为底
![y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x](/uploads/image/z/8806039-7-9.jpg?t=y%3Dlog%284%29%281-2x%2Bx%5E2%29+%3Dlog%282%29%5B%281-x%29%5E2%5D+%2Flog%282%29%284%29+%3D2%5Blog%282%29%7C1-x)
因为:(1-2x+x^2)=(1-x)^2
所以:log(4)(1-2x+x^2)=log(4)[(1-x)^2]
对其使用换底公式(log(a)b=[log(c)b]/[log(c)a]),将以4为底,换为以2为底,就有:
log(4)(1-2x+x^2)=log(4)[(1-x)^2]=log(2)[(1-x)^2]/log(2)4
然后对其进行计算,就有:
log(4)(1-2x+x^2)=log(4)[(1-x)^2]=log(2)[(1-x)^2]/log(2)4=2log(2)(1-x)/2
明白了吗?
继续计算,就有:2log(2)(1-x)/2=log(2)(1-x)
所以:log(4)(1-2x+x^2)=log(4)[(1-x)^2]
对其使用换底公式(log(a)b=[log(c)b]/[log(c)a]),将以4为底,换为以2为底,就有:
log(4)(1-2x+x^2)=log(4)[(1-x)^2]=log(2)[(1-x)^2]/log(2)4
然后对其进行计算,就有:
log(4)(1-2x+x^2)=log(4)[(1-x)^2]=log(2)[(1-x)^2]/log(2)4=2log(2)(1-x)/2
明白了吗?
继续计算,就有:2log(2)(1-x)/2=log(2)(1-x)
y=log(4)(1-2x+x^2) =log(2)[(1-x)^2] /log(2)(4) =2[log(2)|1-x
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