Define $$\ln\frac{c_t}{c}=\hat{c_t}$$ which means % deviation from steady state.
Then $$\frac{1}{c_t +\alpha g_t}=\lambda_t$$ is equivalent as
$$\frac{1}{c e^{\hat{c_t}} + \alpha g e^{\hat{g_t}}}=\lambda e^{\hat{\lambda_t}}$$
$$c e^{\hat{c_t}} + \alpha g e^{\hat{g_t}}=\frac{1}{\lambda} e^{\hat{-\lambda_t}}$$
$$-\frac{c}{c +\alpha g} \hat{c_t} - \frac{\alpha g}{c+\alpha g}\hat{g_t}\approx \hat{\lambda_t}$$
How to derive the last equation with approximation?

Taking a Taylor expansion of $$e^\hat x$$ gives around $$0$$ gives: $$e^\hat x \approx e^0 + e^0 \hat x = 1 + \hat x.$$ Substitution gives: \begin{align*} &(c + \alpha g) + c \hat c_t + \alpha g \hat g_t \approx \frac{1}{\lambda}(1 - \hat \lambda_t),\\ \iff & \lambda(c + \alpha g) + \lambda c \hat c_t + \lambda \alpha g \hat g_t \approx 1 - \hat \lambda_t,\\ \iff & 1 + \frac{c}{c + \alpha g} \hat c_t + \frac{\alpha g}{c + \alpha g} \hat g_t \approx 1 - \hat \lambda_t,\\ \iff & -\frac{c}{c + \alpha g} \hat c_t- \frac{\alpha g}{c + \alpha g}\hat g_t \approx \hat \lambda_t \end{align*}
• Why the $\frac{1}{\lambda(c+\alpha g)}$ term doesn't affect RHS? Oct 16, 2021 at 11:20
• $\lambda = \frac{1}{c + \alpha g}$ so $\frac{1}{\lambda(c + \alpha g)} = 1$.