I am trying to solve a problem that asks to log linearise following Euler equation of the New Keynesian model:
$$C^{-\sigma}_t=\beta E_tC^{-\sigma}_{t+1}(1+i_t)/(1+\pi_{t+1}).$$
The solution is given as: $$\tilde{C}_t=\beta E_t\tilde{C}-\frac{1}{\sigma}(i_t-E_t\tilde{C}_{t+1}-\rho).$$
And also there are following relationships in the steady state given $1=\beta(1+r)$, $\beta=(1+\rho)^{-1}$ and $ln1=ln \beta+r$ what leads to $r=-ln\beta=\rho$.
Can anyone explain to me how the log linearization is done? I know that the first step would be to take the logarithm of everything (and subtract the logarithm of the constant steady state parameter) or to use Taylor approximation. However I have a problem with the $(1+i_t)$ and $(1+\pi_{t+1})$ terms. Furthermore why does $1=\beta(1+r)$ hold in steady state? Note that $i$ denotes nominal interest rate.
Thank you.
UPDATE: So below I will give my solution path. I am not sure though whether it is appropriate to do it like this: $$C^{-\sigma}_t=\beta E_tC^{-\sigma}_{t+1}(1+i_t)/(1+\pi_{t+1})$$ $$lnC^{-\sigma}_t=ln\beta+lnE_t(C^{-\sigma}_{t+1})+ln(1+i_t)-ln(1+\pi_{t+1})\qquad (1)$$ Subtracting (1) with $$lnC^{-\sigma}=ln\beta+ln(C^{-\sigma})+ln(1+i)-ln(1+\pi)$$yields $$\tilde{C}_t=E_t(\tilde{C}_{t+1})-\frac{1}{\sigma}(\tilde{(1+i_t)}-\tilde{(1+\pi_{t+1})})$$ and since $\tilde{1+i_t}=ln(1+i_t)-ln(1+i)=i_t-i$ and $\tilde{1+\pi_{t+1}}=\pi_{t+1}-\pi$: $$\tilde{C}_t=E_t(\tilde{C}_{t+1})-\frac{1}{\sigma}(i_t-i+E\pi_{t+1}-\pi)\qquad(2)$$ $i$ is the nominal interest rate in steady state and can be defined as $i=r+\pi$ and $r=\rho$: $$\tilde{C}_t=E_t(\tilde{C}_{t+1})-\frac{1}{\sigma}(i_t-\rho+\pi-E\pi_{t+1}-\pi)\qquad(3)$$
Leading us to the final equation: $$\tilde{C}_t=\beta E_t\tilde{C}-\frac{1}{\sigma}(i_t-E_t\tilde{C}_{t+1}-\rho).$$