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I want to solve my modified solow equation for k_t to get the steady state. But since I also inclueded the marginal product for the price of Energy its pretty difficult for me to find a solution.

here is the equation

$$ \frac{\delta \cdot k_t}{s \cdot A \cdot k_t^\alpha} = \left(\frac{P_e}{\gamma \cdot A \cdot k_t^\alpha}\right)^{\frac{\gamma}{\gamma-1}} $$

I want to solve for $k_t$

Here is the initial equation.

$$ 0 = s \cdot A \cdot k_t^\alpha \cdot \left(\frac{P_e}{\gamma \cdot A \cdot k_t^\alpha}\right)^{\frac{\gamma}{\gamma-1}} - \delta \cdot k_t $$

I would be really thankful if someone could help me or give me a hint

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1 Answer 1

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$$ \begin{eqnarray} \frac{\delta \cdot k_t}{s \cdot A \cdot k_t^\alpha} & = & \left(\frac{P_e}{\gamma \cdot A \cdot k_t^\alpha}\right)^{\frac{\gamma}{\gamma-1}} \\ \frac{\delta}{s \cdot A} \cdot k_t^{1-\alpha} & = & \left(\frac{P_e}{\gamma \cdot A}\right)^{\frac{\gamma}{\gamma-1}} \cdot k_t^{-\frac{\gamma}{\gamma-1}\alpha} \\ k_t^{1-\alpha} \cdot k_t^{\frac{\gamma}{\gamma-1}\alpha} & = & \frac{s \cdot A}{\delta} \cdot \left(\frac{P_e}{\gamma \cdot A}\right)^{\frac{\gamma}{\gamma-1}} \\ k_t^{1-\frac{\alpha}{\gamma-1}} & = & \frac{s \cdot A}{\delta} \cdot \left(\frac{P_e}{\gamma \cdot A}\right)^{\frac{\gamma}{\gamma-1}} \end{eqnarray} $$ And then you just substitute the parameters and take the right hand side to the proper power.

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