簡約 2ln(2)+9/5 ln(4)+1/5 ln(-1)

解

簡約

2 ln (2) + \frac{9}{5} ln (4) + \frac{1}{5} ln (- 1)

ln (- 32 \cdot 2^{\frac{3}{5}})

解答ステップ

2 ln (2) + \frac{9}{5} ln (4) + \frac{1}{5} ln (- 1)

対数の規則を適用する:

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

\frac{9}{5} ln (4) = ln (4^{\frac{9}{5}})

= 2 ln (2) + ln (4^{\frac{9}{5}}) + \frac{1}{5} ln (- 1)

対数の規則を適用する:

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

\frac{1}{5} ln (- 1) = ln ((- 1)^{\frac{1}{5}})

= 2 ln (2) + ln (4^{\frac{9}{5}}) + ln ((- 1)^{\frac{1}{5}})

対数の規則を適用する:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (4^{\frac{9}{5}}) + ln ((- 1)^{\frac{1}{5}}) = ln (4^{\frac{9}{5}} (- 1)^{\frac{1}{5}})

= 2 ln (2) + ln (4^{\frac{9}{5}} (- 1)^{\frac{1}{5}})

4^{\frac{9}{5}} (- 1)^{\frac{1}{5}} = - 4 \cdot 4^{\frac{4}{5}}

= 2 ln (2) + ln (- 4 \cdot 4^{\frac{4}{5}})

対数の規則を適用する:

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

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

= ln (2^{2}) + ln (- 4 \cdot 4^{\frac{4}{5}})

対数の規則を適用する:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

ln (2^{2}) + ln (- 4 \cdot 4^{\frac{4}{5}}) = ln (2^{2} (- 4 \cdot 4^{\frac{4}{5}}))

= ln (2^{2} (- 4 \cdot 4^{\frac{4}{5}}))

2^{2} (- 4 \cdot 4^{\frac{4}{5}}) = - 32 \cdot 2^{\frac{3}{5}}

= ln (- 32 \cdot 2^{\frac{3}{5}})