簡約 1/4 x^2(2ln^2(3x)-2ln(3x)+1)

解

簡約

\frac{1}{4} x^{2} (2 ln^{2} (3 x) - 2 ln (3 x) + 1)

\frac{x ^{2}}{4}

解答ステップ

\frac{1}{4} x^{2} (2 ln^{2} (3 x) - 2 ln (3 x) + 1)

対数の規則を適用する:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln^{2} (3 x) = ln^{2} ((3 x)^{2})

= \frac{1}{4} x^{2} (ln^{2} ((3 x)^{2}) - 2 ln (3 x) + 1)

対数の規則を適用する:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln (3 x) = ln ((3 x)^{2})

= \frac{1}{4} x^{2} (ln^{2} ((3 x)^{2}) - ln ((3 x)^{2}) + 1)

対数の規則を適用する:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln^{2} ((3 x)^{2}) - ln ((3 x)^{2}) = ln^{2} (\frac{( 3 x ) ^{2}}{( 3 x ) ^{2}})

= \frac{1}{4} x^{2} (ln^{2} (\frac{( 3 x ) ^{2}}{( 3 x ) ^{2}}) + 1)

規則を適用

\frac{a}{a} = 1

= \frac{1}{4} x^{2} (ln^{2} (1) + 1)

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= (ln^{2} (1) + 1) \frac{1 \cdot x ^{2}}{4}

乗算:

1 \cdot x^{2} = x^{2}

= (ln^{2} (1) + 1) \frac{x ^{2}}{4}

ln^{2} (1) = 0

= (0 + 1) \frac{x ^{2}}{4}

簡素化

= \frac{x ^{2}}{4}