簡約-log_{10}(2pi)-1/2 (0.1^2+0.3^2)-2.238

解

簡約

- lo g_{10} (2 π) - \frac{1}{2} (0. 1^{2} + 0. 3^{2}) - 2.238

- lo g_{10} (2) - lo g_{10} (π) - \frac{0.1}{2} - 2.238

十進法表記

- 3.08617 \dots

解答ステップ

- lo g_{10} (2 π) - \frac{1}{2} (0. 1^{2} + 0. 3^{2}) - 2.238

対数の規則を適用する:

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

lo g_{10} (2 π) = lo g_{10} (2) + lo g_{10} (π)

= - (lo g_{10} (2) + lo g_{10} (π)) - \frac{1}{2} (0. 1^{2} + 0. 3^{2}) - 2.238

\frac{1}{2} (0. 1^{2} + 0. 3^{2}) = 0.1 \cdot \frac{1}{2}

= - (lo g_{10} (2) + lo g_{10} (π)) - 0.1 \cdot \frac{1}{2} - 2.238

\frac{1}{2} \cdot 0.1 = \frac{0.1}{2}

= - (lo g_{10} (2) + lo g_{10} (π)) - \frac{0.1}{2} - 2.238

- (lo g_{10} (2) + lo g_{10} (π)) : - lo g_{10} (2) - lo g_{10} (π)

= - lo g_{10} (2) - lo g_{10} (π) - \frac{0.1}{2} - 2.238