簡約 24-12ln(1+e^2)+12ln(2)

解

簡約

24 - 12 ln (1 + e^{2}) + 12 ln (2)

24 + ln (\frac{4096}{( 1 + e ^{2} ) ^{12}})

十進法表記

6.79463 \dots

解答ステップ

24 - 12 ln (1 + e^{2}) + 12 ln (2)

対数の規則を適用する:

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

12 ln (e^{2} + 1) = ln ((e^{2} + 1)^{12})

= 24 - ln ((e^{2} + 1)^{12}) + 12 ln (2)

対数の規則を適用する:

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

12 ln (2) = ln (2^{12})

= 24 - ln ((e^{2} + 1)^{12}) + ln (2^{12})

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^{12}) - ln ((1 + e^{2})^{12}) = ln (\frac{2 ^{12}}{( e ^{2} + 1 ) ^{12}})

= 24 + ln (\frac{2 ^{12}}{( e ^{2} + 1 ) ^{12}})

2^{12} = 4096

= 24 + ln (\frac{4096}{( 1 + e ^{2} ) ^{12}})