Question

Can someone explain how I'm supposed to do this please?

"use the information below to answer the following questions, you buy a commemorative coin for $20.00. the value of the coin increases at a rate of 3.5% per year."
-how much will the coin be worth in 15 years?
- after how many years will the coin have doubled in value
-radioactive iodine is used to determine the health of the thyroid gland. it decays according to the equation y=ae^-0.0856t, where t is in days. find the half-life of this substance.

Answer 1

See answers below

Explanation:

Given: `$20` value increase at rate `= (3.5%)` per year
`" "`radioactive decay `y = ae^(-.0856t)`

Calculate value after 15 years:

Use the equation `P(t) = P_o (1+r)^t`; where

`P(t) =`the amount of money you make,
`P_o` = initial investment, `r = %/100`
`t =`time in years

note: increasing is `(1+r)`; decreasing is `(1-r)`

After 15 years: `P(t) = 20 (1+.035)^15 ~~ $33.51`

Calculate Doubling time (`40` is double of `20`):

`40 = 20 (1.035)^t`

`40/20 = 1.035^t`

`2 = 1.035^t`

Log both sides (either log or ln can be found on a calculator):

`log 2 = log 1.035^t`

Use the log power rule `log x^c = c log x`:

`log 2 = t log 1.035`

`(log 2)/(log 1.035) = t (log 1.035)/(log 1.035)`

`t = (log 2)/(log 1.035) ~~ 20.15` years

Find the half-life of radioactive decay:

`y = ae^(-.0856t)`; `" "a =`the amount of substance initially

half-life is when half of the substance is left: `1/2a`

`1/2a = a e^(-.0856t)`

`1/2 = e^(-.0856t)`

log both sides. Since `log_e = ln`, use this:

`ln (1/2) = ln e^(-.0856t)`

Use the property `ln e^x = x`

`ln (.5) = -.0856t`

`(ln (.5))/(-.0856) = t`

`t ~~8.1` days