簡約 ln(c^1)+0.8*ln(c^2)

解

簡約

ln (c^{1}) + 0.8 \cdot ln (c^{2})

ln (c^{2.6})

解答ステップ

ln (c^{1}) + 0.8 ln (c^{2})

対数の規則を適用する:

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

0.8 ln (c^{2}) = ln ((c^{2})^{0.8})

= ln (c^{1}) + ln ((c^{2})^{0.8})

対数の規則を適用する:

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

ln (c^{1}) + ln ((c^{2})^{0.8}) = ln (c^{1} (c^{2})^{0.8})

= ln (c^{1} (c^{2})^{0.8})

簡素化

c^{1} (c^{2})^{0.8} : c^{2.6}

= ln (c^{2.6})