簡約 1/9log_{2}(1/9)+8/9log_{2}(8/9)

解

簡約

\frac{1}{9} \cdot lo g_{2} (\frac{1}{9}) + \frac{8}{9} \cdot lo g_{2} (\frac{8}{9})

lo g_{2} (\frac{4 3 2 ^{2}}{9 9 ^{8} 9 3 ^{2}})

十進法表記

- 0.50325 \dots

解答ステップ

\frac{1}{9} lo g_{2} (\frac{1}{9}) + \frac{8}{9} lo g_{2} (\frac{8}{9})

\frac{1}{9} lo g_{2} (\frac{1}{9}) = - \frac{2}{9} lo g_{2} (3)

= - \frac{2}{9} lo g_{2} (3) + \frac{8}{9} lo g_{2} (\frac{8}{9})

- \frac{2}{9} lo g_{2} (3) = - lo g_{2} (3^{\frac{2}{9}})

\frac{8}{9} lo g_{2} (\frac{8}{9}) = lo g_{2} ((\frac{8}{9})^{\frac{8}{9}})

= - lo g_{2} (3^{\frac{2}{9}}) + lo g_{2} ((\frac{8}{9})^{\frac{8}{9}})

対数の規則を適用する:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{2} ((\frac{8}{9})^{\frac{8}{9}}) - lo g_{2} (3^{\frac{2}{9}}) = lo g_{2} \frac{( \frac{8}{9} ) ^{\frac{8}{9}}}{3 ^{\frac{2}{9}}}

= lo g_{2} \frac{( \frac{8}{9} ) ^{\frac{8}{9}}}{3 ^{\frac{2}{9}}}

\frac{( \frac{8}{9} ) ^{\frac{8}{9}}}{3 ^{\frac{2}{9}}} = \frac{4 \cdot 2 ^{\frac{2}{3}}}{9 ^{\frac{8}{9}} \cdot 3 ^{\frac{2}{9}}}

= lo g_{2} (\frac{4 \cdot 2 ^{\frac{2}{3}}}{9 ^{\frac{8}{9}} \cdot 3 ^{\frac{2}{9}}})

2^{\frac{2}{3}} = 3 2^{2}

= lo g_{2} (\frac{4 3 2 ^{2}}{9 9 ^{8} 9 3 ^{2}})

9^{\frac{8}{9}} = 9 9^{8}

= lo g_{2} (\frac{4 3 2 ^{2}}{9 9 ^{8} 9 3 ^{2}})

3^{\frac{2}{9}} = 9 3^{2}

= lo g_{2} (\frac{4 3 2 ^{2}}{9 9 ^{8} 9 3 ^{2}})