What must be invested today, to be worth $20,000 in 10 years, if it is compounded yearly at 8%?

Question

1.What must be invested today, to be worth $20,000 in 10 years, if it is compounded yearly at 8%?

2. Write as a sum or difference of simple (single symbol argument) logs: Log [(AB)/(CD)]

3. The half-life of carbon 14 is 5730 years. How old is a bone containing 20% of its original carbon 14?

Answer 1

As for the first part...

Yes, the amount invested will be $9,263.87 to get $20,ooo compounded yearly at 8% for 10 years

 

A = P(1+r/n)^nt      A=20000    r=.08    n=1    t = 10

$20,000 = P(1+.08)¹⁰

P = 20000/(1.08)¹⁰

P = $9,263.87

 

2)  Log [(AB)/(CD)]

     = Log(AB) - Log(CD)

     = Log(A) + Log(B) - (Log(C) + Log(D))

     = Log(A) + Log(B) - Log(C) - Log(D)

 

3) The half life formula is:

   N = N₀(1/2)^t/t_1/2

 

N = future amount = 20% of starting amount or .2x

N₀ = starting amount = x

t = number of years

t_1/2 = half life = 5730

 

.2x = x(1/2)^t/5730   divide both sides by x

.2 = (.5)^t/5730         take log of both sides

log(.2) = log(.5^t/5730)

log(.2) = (t/5730)[log(.5)]

5730log(.2) = t[log(.5)]

[(5730log(.2)]/[log(.5)] = t

 

t = 13304.65 years

 

rounded to the nearest year the carbon14 is 13305 years old

 

    

Answer 2

To be exact it should be 1 cent greater, because of the  rounding :-)

Answer 3

You would have to invest the sum of $9,263.86 dollars to get $20,ooo compounded yearly at 8% for 10 years