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人気のある代数 >

簡約-0.38*log_{2}(0.38)-0.62log_{2}(0.62)

解

簡約

- 0.38 \cdot lo g_{2} (0.38) - 0.62 lo g_{2} (0.62)

解

- lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{25}) + 1

+1

十進法表記

0.95804 \dots

解答ステップ

- 0.38 lo g_{2} (0.38) - 0.62 lo g_{2} (0.62)

- 0.38 lo g_{2} (0.38) = - 0.38 lo g_{2} (\frac{19}{50})

lo g_{2} (0.62) = lo g_{2} (\frac{31}{50})

= - 0.38 lo g_{2} (\frac{19}{50}) - 0.62 lo g_{2} (\frac{31}{50})

- 0.38 lo g_{2} (\frac{19}{50}) = - lo g_{2} ((\frac{19}{50})^{0.38})

- 0.62 lo g_{2} (\frac{31}{50}) = - lo g_{2} ((\frac{31}{50})^{0.62})

= - lo g_{2} ((\frac{19}{50})^{0.38}) - lo g_{2} ((\frac{31}{50})^{0.62})

対数の規則を適用する:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

- lo g_{2} ((\frac{31}{50})^{0.62}) - lo g_{2} ((\frac{19}{50})^{0.38}) = lo g_{2} ((\frac{31}{50})^{0.62} (\frac{19}{50})^{0.38})

= - lo g_{2} ((\frac{31}{50})^{0.62} (\frac{19}{50})^{0.38})

簡素化

(\frac{31}{50})^{0.62} (\frac{19}{50})^{0.38} : \frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{50}

= - lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{50})

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_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{50}) = lo g_{2} (3 1^{0.62} \cdot 1 9^{0.38}) - lo g_{2} (50)

= - (lo g_{2} (3 1^{0.62} \cdot 1 9^{0.38}) - lo g_{2} (50))

lo g_{2} (3 1^{0.62} \cdot 1 9^{0.38}) = 0.62 lo g_{2} (31) + 0.38 lo g_{2} (19)

= - ((0.62 lo g_{2} (31) + 0.38 lo g_{2} (19)) - lo g_{2} (50))

規則を適用する:

(a) = a

(0.62 lo g_{2} (31) + 0.38 lo g_{2} (19)) = 0.62 lo g_{2} (31) + 0.38 lo g_{2} (19)

= - (0.62 lo g_{2} (31) + 0.38 lo g_{2} (19) - lo g_{2} (50))

lo g_{2} (50) = 1 + 2 lo g_{2} (5)

= - (0.62 lo g_{2} (31) + 0.38 lo g_{2} (19) - (1 + 2 lo g_{2} (5)))

分配法則を適用する:

- (a + b) = - a - b

- (1 + 2 lo g_{2} (5)) = - 1 - 2 lo g_{2} (5)

= - (0.62 lo g_{2} (31) + 0.38 lo g_{2} (19) - 1 - 2 lo g_{2} (5))

0.62 lo g_{2} (31) = lo g_{2} (3 1^{0.62})

0.38 lo g_{2} (19) = lo g_{2} (1 9^{0.38})

- 2 lo g_{2} (5) = - lo g_{2} (5^{2})

= - (lo g_{2} (3 1^{0.62}) + lo g_{2} (1 9^{0.38}) - 1 - lo g_{2} (5^{2}))

lo g_{2} (3 1^{0.62}) + lo g_{2} (1 9^{0.38}) - lo g_{2} (5^{2}) = lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{25})

= - (lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{25}) - 1)

分配法則を適用する:

- (a - b) = - a + b

- (lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{25}) - 1) = - lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{25}) + 1

= - lo g_{2} (\frac{3 1 ^{0.62} \cdot 1 9 ^{0.38}}{25}) + 1

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