簡約 0.5ln(x^2+1)-4ln(0.5)-0.5(ln(x+4)+ln(x))

解

簡約

0.5 ln (x^{2} + 1) - 4 ln (0.5) - 0.5 (ln (x + 4) + ln (x))

ln (\frac{16 ( x ^{2} + 1 ) ^{0.5}}{( ( x + 4 ) x ) ^{0.5}})

解答ステップ

0.5 ln (x^{2} + 1) - 4 ln (0.5) - 0.5 (ln (x + 4) + ln (x))

対数の規則を適用する:

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

0.5 ln (x^{2} + 1) = ln ((x^{2} + 1)^{0.5})

= ln ((x^{2} + 1)^{0.5}) - 4 ln (0.5) - 0.5 (ln (x + 4) + ln (x))

対数の規則を適用する:

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

4 ln (0.5) = ln (0. 5^{4})

= ln ((x^{2} + 1)^{0.5}) - ln (0. 5^{4}) - 0.5 (ln (x + 4) + ln (x))

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 ((x^{2} + 1)^{0.5}) - ln (0. 5^{4}) = ln (\frac{( x ^{2} + 1 ) ^{0.5}}{0. 5 ^{4}})

= ln (\frac{( x ^{2} + 1 ) ^{0.5}}{0. 5 ^{4}}) - 0.5 (ln (x + 4) + ln (x))

簡素化

\frac{( x ^{2} + 1 ) ^{0.5}}{0. 5 ^{4}} : 16 (x^{2} + 1)^{0.5}

= ln (16 (x^{2} + 1)^{0.5}) - 0.5 (ln (x + 4) + ln (x))

対数の規則を適用する:

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

ln (x + 4) + ln (x) = ln (x (x + 4))

= ln (16 (x^{2} + 1)^{0.5}) - 0.5 ln ((x + 4) x)

対数の規則を適用する:

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

0.5 ln (x (x + 4)) = ln ((x (x + 4))^{0.5})

= ln (16 (x^{2} + 1)^{0.5}) - ln ((x (x + 4))^{0.5})

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 (16 (x^{2} + 1)^{0.5}) - ln ((x (x + 4))^{0.5}) = ln (\frac{16 ( x ^{2} + 1 ) ^{0.5}}{( x ( x + 4 ) ) ^{0.5}})

= ln (\frac{16 ( x ^{2} + 1 ) ^{0.5}}{( ( x + 4 ) x ) ^{0.5}})