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I am trying to get a good understanding of the steps involved in solving the dual of a maximization problem, namely cost minimization. At some point (last two steps), the author ends up with the following function for marginal cost growth: $$\Delta mc=\left [\frac{WN}{C(\cdot )} \right ]\Delta w+\left [1-\frac{WN}{C(\cdot )} \right ]\Delta r -\Delta e$$ Where the relation between price and marginal cost is given by: $$(1-B)P=MC=\frac{G(W,R)}{E}$$ From this latter equation, the difference between the change in price and a weighted average of changes in factor prices, the dual or price-based Solow residual is defined as: $$\alpha \Delta w +(1-\alpha)\Delta r-\Delta p=-B(\Delta p-\Delta r)+(1-B)\Delta e$$ Where the weights for the factor prices are the wage share in output for wages and its complement for capital costs ($\alpha$) and ($1-\alpha$).

I can't figure out how you get to this last step. I tried differentiating the marginal cost equation and substituting in the second one, but I'm missing some terms. The problem comes from the following paper: "Can Imperfect Competition Explain the Difference between Primal and Dual Productivity Measures? Estimates for U.S. Manufacturing" (1995).

*$\Delta x$ is the log difference of the variable X.

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Equation (4b) in the paper gives: $$ \Delta mc_t = \frac{E_t N_t W_t}{Y_t G(.)}\Delta w_t + \frac{E_t K_t R_t}{Y_t G(.)} \Delta r_t - \Delta e_t. \tag{I} $$ Next we also have equation (5) in the paper: $$ (1-B)P_t = MC_T = \frac{G(.)}{E_t}. \tag{II} $$ Now substitute $(II)$ into $(I)$ to get: $$ \Delta mc_t = \frac{N_t W_t}{P_t Y_t (1-B)}\Delta w_t + \frac{K_t R_t}{P_t Y_t (1-B)} \Delta r_t - \Delta e_t. \tag{III} $$ Now, we define the wage cost over the total revenue to be $\alpha_t$. $$ \alpha_t = \frac{N_t W_t}{P_t Y_t}. \tag{IV} $$ Also, by the CRS assumption, we have that $\frac{N_t W_t}{C(.)} + \frac{R_t K_t}{C(.)} = 1$. Dividing this by $P_t Y_t/C(.)$ and using $C(.) = G(.)Y_t/E_t$ as equation (2) in the paper, we get: $$ \frac{N_t W_t}{P_t Y_t} + \frac{R_t K_t}{P_t Y_t} = \frac{C(.)}{P_t Y_t} = \frac{G(.)Y_t}{E_t P_t Y_t} = \frac{G(.)}{E_t P_t} = (1-B) \tag{V} $$ The last equality comes from comes from $(II)$ Using $(IV)$ in $(V)$ gives: $$ \frac{R_t K_t}{P_t Y_t} = (1-B) - \alpha_t. \tag{VI} $$ Now substitute $(IV)$ and $(VI)$ into $(III)$ to get: $$ \Delta mc_t = \frac{\alpha_t}{(1-B)} \Delta w_t + \frac{1 - \alpha_t - B}{(1-B)}\Delta r_t - \Delta e_t. \tag{VII} $$ Also log differentiating the first part of $(II)$ gives $\Delta p_t = \Delta mc_t$ so substituting this in $(VII)$ (and multiplying both sides by $(1-B)$ finally produces: $$ \Delta p_t (1-B) = \alpha_t \Delta w_t + (1 - \alpha_t - B) \Delta r_t - \Delta e_t. \tag{VIII} $$ The expression that you are looking for is a simple rearrangement of $(VIII)$.

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