簡約-0.6log_{2}(0.6)-0.3log_{2}(0.3)

解

簡約

- 0.6 \cdot lo g_{2} (0.6) - 0.3 \cdot lo g_{2} (0.3)

- lo g_{2} (0.51289 \dots)

十進法表記

0.96326 \dots

解答ステップ

- 0.6 lo g_{2} (0.6) - 0.3 lo g_{2} (0.3)

対数の規則を適用する:

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

0.6 lo g_{2} (0.6) = lo g_{2} (0. 6^{0.6})

= - lo g_{2} (0. 6^{0.6}) - 0.3 lo g_{2} (0.3)

対数の規則を適用する:

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

0.3 lo g_{2} (0.3) = lo g_{2} (0. 3^{0.3})

= - lo g_{2} (0. 6^{0.6}) - lo g_{2} (0. 3^{0.3})

- lo g_{2} (0. 6^{0.6}) - lo g_{2} (0. 3^{0.3})

を書き換え

- (lo g_{2} (0. 6^{0.6}) + lo g_{2} (0. 3^{0.3}))

= - (lo g_{2} (0. 6^{0.6}) + lo g_{2} (0. 3^{0.3}))

対数の規則を適用する:

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

lo g_{2} (0. 6^{0.6}) + lo g_{2} (0. 3^{0.3}) = lo g_{2} (0. 6^{0.6} \cdot 0. 3^{0.3})

= - lo g_{2} (0. 6^{0.6} \cdot 0. 3^{0.3})

0. 6^{0.6} \cdot 0. 3^{0.3} = 0.51289 \dots

= - lo g_{2} (0.51289 \dots)