簡約 d+θln(k)+γln(ω)-ln(1/(1+βθ)k^αω)-βd

解

簡約

d + θ ln (k) + γ ln (ω) - ln (\frac{1}{1 + βθ} k^{α} ω) - β d

d + θ ln (k) + γ ln (ω) + ln (1 + θβ) - α ln (k) - ln (ω) - d β

解答ステップ

d + θ ln (k) + γ ln (ω) - ln (\frac{1}{1 + βθ} k^{α} ω) - β d

対数の規則を適用する:

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

ln (\frac{1}{1 + βθ} k^{α} ω) = ln (\frac{1}{1 + βθ}) + ln (k^{α}) + ln (ω)

= d + θ ln (k) + γ ln (ω) - (ln (k^{α}) + ln (ω) + ln (\frac{1}{θβ + 1})) - d β

対数の規則を適用する:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln (k^{α}) = α ln (k)

= d + θ ln (k) + γ ln (ω) - (ln (\frac{1}{1 + βθ}) + α ln (k) + ln (ω)) - β d

対数の規則を適用する:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= d + θ ln (k) + γ ln (ω) - (α ln (k) + ln (ω) - ln (θβ + 1)) - d β

= d + θ ln (k) + γ ln (ω) - (- ln (1 + θβ) + α ln (k) + ln (ω)) - d β

- (- ln (1 + βθ) + α ln (k) + ln (ω)) : ln (1 + βθ) - α ln (k) - ln (ω)

= d + θ ln (k) + γ ln (ω) + ln (1 + βθ) - α ln (k) - ln (ω) - β d

= d + θ ln (k) + γ ln (ω) + ln (1 + θβ) - α ln (k) - ln (ω) - d β