A = Pe^(rt)

1/6 = e^(r*4) .. -ln(6)/4 = r


Pe^(-ln(6)/4 * 20) = how much there after 20 years

there now = original - lost

lost = original - lost = P(1 - e^(ln(5/6)/4 * 20)

percentage : lost/original 

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A = 52e^(-rt)
13 = 52 e^(-5r)
13/52 = e^(-5r) => line 30									



A = Pe^(-rt)

52 = Pe^(-r3)	... P = 52/(e^(-r3)) ... P = 52 e^(3r)

13 = Pe^(-r8)   ... P = 13/(e^(-r8)) ... P = 13 e^(8r)

52/13 = e^(8r) / e^(3r) = e^(8r) * e^(-3r) = e^(5r) 

ln(52/13) = 5r

ln(52/13)/5 = r

u = log(3); v = log(5);

2u = 2log(3) = log(9)
2v = 2log(5) = log(25)

log(.9) = log(9/10) = log(9) - log(10)  = -1 + 2u 

log(250) = log(25*10) = log(25) + log(10) = 2
.5*log(1000) = .5*log(5*2*100) = .5(v + log(200)) = .5v * .5 log(2*100) =

log(1000) = log(100*10) = log(100) + 1 = log(20) + v + 1 

.5(3u + 2v) = 
3u + 2v = 3log(3) + 2log(5) = log(27) + log(25)