How to calculate GDP increase using Cobb-Douglas production function? I use this economics stack exchange for the first time. I have a question regarding the Cobb-Douglas production function. The question is this: GDP in an economy is growing at 3% a year in real terms. Population is constant. The government decides to allow a significant increase in immigration so that the population (and the workforce) starts to grow by 1% a year. Output is produced in the economy according to a Cobb-Douglas production function. The share of labour income in GDP is 70%. How much higher will GDP be as a result of the new immigration policy after 20 years? How much higher will GDP per capita be?

I am not be able to understand how to use the function in this problem. Sorry if I put a question in a wrong way. Thank you.
I think this approach might be useful.The Cobb douglas production function with constant returns to scale is 
\begin{equation}
Q=K^{\alpha}L^{1-\alpha}
\end{equation}
For simplicity, we use logarithms for the rates of change of its determinants
\begin{equation}
ln(Q)=\alpha ln(K)+(1-\alpha)ln(L)
\end{equation}

Differentiating with respect to time on both sides of the equation

\begin{equation}
\frac{1}{Q}\frac{dQ}{dt}=\alpha\frac{1}{K}\frac{dK}{dt}+(1-\alpha)\frac{1}{L}\frac{dL}{dt} 
\end{equation}

\begin{equation}
\begin{array}{l}
 \text{Growth rate of GDP:}\frac{1}{Q}\frac{dQ}{dt}=G_{Q} \\
\text{Growth rate of Capital:}\frac{1}{K}\frac{dK}{dt}=G_{K} \\
\text{Growth rate of Labor:}\frac{1}{L}\frac{dL}{dt}=G_{L}
\end{array}
\end{equation}
\begin{equation}
G_{Q}=\alpha G_{K}+(1-\alpha)G_{L}
\end{equation}
Where $\alpha$ is the capital share in the gross domestic product. In your case $(1-\alpha)=70\%$, hence $\alpha=30\%$ 

Thus, 
$G_{k}=0\%$ (The capital in this case does not change.), $G_{L}=1\%$ 
Substituting in the original equation
\begin{equation}
G_{Q}=0.7\cdot 1\%=0.7\%
\end{equation}
for further growth rate.

Adding to the previous growth rate ($3\%$), we obtain
$3\%+0.7\%=3.7\%$
To find the new GDP after 20 years.
\begin{equation}
 (1+0.037)^{20}=2.068117 
\end{equation}