簡約 2ln(2)+5ln(x)-1/2 ln(x+9)

解

簡約

2 ln (2) + 5 ln (x) - \frac{1}{2} ln (x + 9)

ln (\frac{4 x ^{5} x + 9}{x + 9})

解答ステップ

2 ln (2) + 5 ln (x) - \frac{1}{2} ln (x + 9)

対数の規則を適用する:

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

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

= ln (2^{2}) + 5 ln (x) - \frac{1}{2} ln (x + 9)

対数の規則を適用する:

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

5 ln (x) = ln (x^{5})

= ln (2^{2}) + ln (x^{5}) - \frac{1}{2} ln (x + 9)

対数の規則を適用する:

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

ln (2^{2}) + ln (x^{5}) = ln (2^{2} x^{5})

= ln (2^{2} x^{5}) - \frac{1}{2} ln (x + 9)

2^{2} = 4

= ln (4 x^{5}) - \frac{1}{2} ln (x + 9)

対数の規則を適用する:

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

\frac{1}{2} ln (x + 9) = ln ((x + 9)^{\frac{1}{2}})

= ln (4 x^{5}) - ln ((x + 9)^{\frac{1}{2}})

対数の規則を適用する:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln (4 x^{5}) - ln ((x + 9)^{\frac{1}{2}}) = ln (\frac{4 x ^{5}}{( x + 9 ) ^{\frac{1}{2}}})

= ln (\frac{4 x ^{5}}{( x + 9 ) ^{\frac{1}{2}}})

簡素化

\frac{4 x ^{5}}{( x + 9 ) ^{\frac{1}{2}}} : \frac{4 x ^{5} x + 9}{x + 9}

= ln (\frac{4 x ^{5} x + 9}{x + 9})