簡約 log_{10}(((x+1)(x-8))/((x-1)(x+8)))

解

簡約

lo g_{10} (\frac{( x + 1 ) ( x - 8 )}{( x - 1 ) ( x + 8 )})

lo g_{10} (x + 1) + lo g_{10} (x - 8) - lo g_{10} (x - 1) - lo g_{10} (x + 8)

解答ステップ

lo g_{10} (\frac{( x + 1 ) ( x - 8 )}{( x - 1 ) ( x + 8 )})

対数の規則を適用する:

lo g_{a} (\frac{x}{y}) = lo g_{a} (x) - lo g_{a} (y)

lo g_{10} (\frac{( x + 1 ) ( x - 8 )}{( x - 1 ) ( x + 8 )}) = lo g_{10} ((x + 1) (x - 8)) - lo g_{10} ((x - 1) (x + 8))

= lo g_{10} ((x + 1) (x - 8)) - lo g_{10} ((x - 1) (x + 8))

lo g_{10} ((x + 1) (x - 8)) = lo g_{10} (x + 1) + lo g_{10} (x - 8)

lo g_{10} ((x - 1) (x + 8)) = lo g_{10} (x - 1) + lo g_{10} (x + 8)

= lo g_{10} (x + 1) + lo g_{10} (x - 8) - (lo g_{10} (x - 1) + lo g_{10} (x + 8))

分配法則を適用する:

- (a + b) = - a - b

- (lo g_{10} (x - 1) + lo g_{10} (x + 8)) = - lo g_{10} (x - 1) - lo g_{10} (x + 8)

= lo g_{10} (x + 1) + lo g_{10} (x - 8) - lo g_{10} (x - 1) - lo g_{10} (x + 8)