Derivation of exponential function (minor help)

Question

This is basic maths but I'm just not seeing the connection here. How is 2-3 equation derived?

$$e^{-\gamma(c_{t-1}+K_{t-1}+\nu_{t})}=E_te^{-\gamma(c_{t-1}+K_{t-1}+\nu_{t}+K_t+\nu_{t+1})}$$

$$e^{\gamma K_t}=E_te^{\gamma\nu_{t+1} }$$ $$K_t=\frac{1}{\gamma}log(E_te^{\gamma\nu_{t+1}})$$

Answer 1

Remember $e^{-m}e^{+m}=e^{-m+m}=e^0=1$.

Multiply (1) with $e^{+\gamma(c_{t-1}+K_{t-1}+\nu_{t})} $ on both sides:

$$e^{-\gamma(c_{t-1}+K_{t-1}+\nu_{t})} e^{+\gamma(c_{t-1}+K_{t-1}+\nu_{t})} =E_te^{-\gamma(c_{t-1}+K_{t-1}+\nu_{t}+K_t+\nu_{t+1})} e^{+\gamma(c_{t-1}+K_{t-1}+\nu_{t})} $$

$$e^{\gamma K_t}=E_te^{\gamma\nu_{t+1} } \tag 2$$

Again, from the rules of logarithm:

$\log a^m=m\log a, \log e =1$

Take logarithm with both sides of (2):

We get $${\gamma K_t}\log e =\log (E_te^{\gamma\nu_{t+1} })$$ $$K_t=\frac{1}{\gamma}log(E_te^{\gamma\nu_{t+1}})\tag 3$$

Answer 2

You can cancel out terms in the exponents on different sides of the equation. $c_{t-1}$, $K_t $, and $\nu_t$ appear on both sides, so you can cancel them. More explicitly,

$$e^{-\gamma(c_{t-1}+K_{t-1}+\nu_{t})}=E_te^{-\gamma(c_{t-1}+K_{t-1}+\nu_{t}+K_t+\nu_{t+1})}$$

distribute $-\gamma$

$$e^{-\gamma c_{t-1}-\gamma K_{t-1}-\gamma \nu_{t}}=E_te^{-\gamma c_{t-1}-\gamma K_{t-1}-\gamma \nu_{t}-\gamma K_t-\gamma \nu_{t+1})}$$

multiplication in exponent corresponds to multiplication ($e^{a+b} = e^ae^b$).

$$e^{-\gamma c_{t-1}-\gamma K_{t-1}-\gamma \nu_{t}}=E_t(e^{-\gamma c_{t-1}-\gamma K_{t-1}-\gamma \nu_{t}})(e^{-\gamma K_t-\gamma \nu_{t+1}})$$

Divide both sides by $e^{-\gamma c_{t-1}-\gamma K_{t-1}-\gamma \nu_{t}}$

$$1=E_te^{-\gamma K_t-\gamma \nu_{t+1}}$$ $$1=E_te^{-\gamma K_t}e^{-\gamma \nu_{t+1}}$$

multiply both sides by $e^{\gamma K_t}$ $$e^{\gamma K_t}=E_te^{-\gamma \nu_{t+1}}$$

take the log of both sides

$${\gamma}K_t=log(E_te^{\gamma\nu_{t+1}})$$

divide both sides by $\gamma$

$$K_t=\frac 1 {\gamma}log(E_te^{\gamma\nu_{t+1}})$$