A few exponentical word problems?

Question

1. a) How long would it take for a 1 g sample of Polonium 210 to lose 3/4 of its mass if it's half-life is 140 days?

b) How long will it take before it loses all but 1/128 of its inital mass?

Answers a= 280 b= 980 days

2. At the end of 15 days, one eight of a raidoactive sample of Bismuth 210 remains. What is the half-life of this element?

answer 5 days

can't get these answers

Answer 1

1a. .025 = 0.5^(t/140)
ln 0.25 = t/140 ln 0.5
t = 140 * ln 0.25/ln 0.5 = 280

1b. t = 140 * ln (1/128)/ln 0.5 - 980

1/8 = 1/2^(15/b)
ln 0.125 = 15/b ln 0.5
b = 15 * ln 0.5/ln0.125
= 5

Answer 2

All equations are of the form a = e^(-kt).

1a) .5 = e^(-kt). First we must determine k from this equation. Solving for k=ln(.5)/(-140) = .0049510513.

.25 = e^(-.004951053t). Solving for t = ln(.25)/.0049510513 = 280 days.

1b) 1/128 = e^(-.004951053t). Solving for t= ln(1/128)/.004951053 = 980 days.

2) y=e^(-kt). 1/8 = e^(-k/15). Solving for k=ln(1/8)/(-15) = .1386294361.

.5 = e^(-.1386294361t). Solving for t = ln(.5)/(-.1386294361) = 5 days.

Answer 3

exponential decay can be expressed as

Q(t) = Q(0)e^kt

where Q(t) = quantity present at time

Q(0) = quantity present intially

k = decay constant and t = time

a)

first find out decay constant
here Q(0) = 1 g , Q(140) = 1/2 = 0.5 : and t = 140 dyas

so 0.5 = e^140k

140 k = ln(0.5)

k = ln(0.5)/140 = -4.95*(10^-3)

after losing 3/4 mass, the amount present = 1-3/4 = 1/4 = 0.25g

so 0.25 = e ^(-4.95*10^-3t)

-4.95(10^-3) t = ln(0.25)

t = -ln0.25/4.95(10^-3) = 280 days

b)

here Q(t) = 1/128

so 1/128 = e^(-4.95*10^-3) t

-4.95(10^-3)t = ln(1/128)

t = -ln(1/128)/4.95*10^-3 = 980 dys

2)

here Q(t) = 1/8, t = 15 days and let intial amount Q(0) = x

so 1/8 = xe^(15k)

1/8x = e^15k

15k = ln(1/8x)

k = (1/15)ln(1/8x) ------------------eqn(1)

and for half-life , the equation is 1/2 = xe^kt

1/2x = e^kt

kt = ln(1/2x)

t = (1/k)ln(1/2x)

substitute k value from eqn(1)

= 15/ln(8x)* ln(1/2x)

t = 15ln(1/2x) / 3ln(1/2x) (since ln(8x) = ln[2^3*(x)] = 3ln(2x))

t = 15/3 = 5 days