Let's assume
It started in 2000
so, t=0 in 2000
![P_0=6.4million](https://img.qammunity.org/2019/formulas/mathematics/middle-school/uoxjvnyhv707p2ba6mq4hlraodbzvf4wr9.png)
we can use formula
![P(t)=P_0 e^(rt)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/v2bsfsbbssbs1sqqag24ik799cg8xypbpu.png)
we can plug value
![P(t)=6.4 e^(rt)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4x50nnsfg5mfur99mvgp54kt2x4kraoa8f.png)
In 2005, the same survey was made and the total amount of gamma ray bursts was 7.3 million
so, at t=2005-2000=5
P(t)=7.3 million
we can plug value and then we can solve for r
![7.3=6.4 e^(r*5)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/84zv7oiw20j53o3o18fi7onma1k715r1pl.png)
![r=0.02632](https://img.qammunity.org/2019/formulas/mathematics/middle-school/irc5nv07ht7kdo4r23c947b89gqj8o4dh0.png)
now, we can plug back
![P(t)=6.4 e^(0.02632t)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/y0tldrtes4wabcxdqqxb5ueat89ti6398k.png)
now, we have
P(t)=1 billion =1000 million
so, we can set it and then we can solve for t
![1000=6.4 e^(0.02632t)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ofo8723u8ay128gdhkuaac49t5uyhrjgrl.png)
![t=191.924](https://img.qammunity.org/2019/formulas/mathematics/middle-school/lirg3y5tu5dcx0wbnjiwb86lix1xvb96pc.png)
approximately
![t=192](https://img.qammunity.org/2019/formulas/mathematics/middle-school/81pys9k8xqzjzszfas56rlmfwxif6xo2sj.png)
Year is 2000+192
year is 2192................Answer