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

The solution to d+θln(k)+γln(ω)-ln(1/(1+βθ)k^αω)-βd is d+θln(k)+γln(ω)+ln(1+θβ)-αln(k)-ln(ω)-dβ

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

simplify

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

Solution steps

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

Apply log rule:

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

Apply log rule:

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

Apply log rule:

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

6 lo g_{8} (u) + 6 lo g_{8} (w) - 36 lo g_{8} (v)

- lo g_{10} (5 \cdot 1 0^{8})

ln \frac{( x ( x ^{2} + 1 ) ^{2} )}{2 x ^{3} - 1}

4 \cdot ln (- 1) + ln (4)

lo g_{e} (5 x^{- 6})

What is d+θln(k)+γln(ω)-ln(1/(1+βθ)k^αω)-βd ?
The solution to d+θln(k)+γln(ω)-ln(1/(1+βθ)k^αω)-βd is d+θln(k)+γln(ω)+ln(1+θβ)-αln(k)-ln(ω)-dβ