Let's take it from what you have, Quazars......
We have
1000 = 100000(.8)n
Divide both sides by 100000
1/100 = (.8)n
Take the log of both sides
log(1/100) = log (.8)n
And by a property of logs, we can bring the "n" out front
log(1/100) =n log(.8)
Divide both sides by log (.8)
log(1/100) / log(.8) = n = about 21 wars are required