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


1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.