解下列方程:(1)log2(4-x)-log4(x-1)=1 (2)2lg(2x-1)=lg(-2x+7)+lg(x+1

2个回答

  • log2(4-x)-log4(x-1)=1

    log2(4-x)-log2(x-1)/log2(4)=1

    log2(4-x)-log2(x-1)/2=1

    2log2(4-x)-log2(x-1)=2

    log2(4-x)²-log2(x-1)=2

    log2[(4-x)²/(x-1)]=2

    则(4-x)²/(x-1)=4

    (4-x)²=4(x-1)

    16-8x+x²=4x-4

    20-12x+x²=0

    (x-2)(x-10)=0

    x1=2 x2=10

    2lg(2x-1)=lg(-2x+7)+lg(x+1)

    2lg(2x-1)-lg(-2x+7)-lg(x+1)=0

    lg(2x-1)²-lg(-2x+7)-lg(x+1)=0

    lg{(2x-1)²/[(-2x+7)(x+1)]}=0

    则(2x-1)²/[(-2x+7)(x+1)]=1

    (2x-1)²=(-2x+7)(x+1)

    4x²-4x+1=-2x²-2x+7x+7

    6x²-9x-6=0

    2x²-3x-2=0

    (2x+1)(x-2)=0

    x1=-1/2 x2=2