Question

A radioactive substance decays at a continuous rate of 17% per year, and 250 mg of the substance is present in the year 2009.

Answer 1

Answer:

a) $A(t) = 250 (1-0.17)^t = 250(0.83)^t$

b) $A(t=14) = 250 (0.83)^{14}= 18.408 mg$

c) For this case we want to find when the quantity is below 25 mg, so we can do this:

$25 = 250 (0.83)^t$

We can divide both sided by 250 and we got:

$0.1 = 0.83^t$

Now we can apply natural log on both sides and we got:

$ln(0.1) = t ln (0.83)$

And if we solve for t we got:

$t = \frac{ln(0.1)}{ln(0.83)}= 12.358 years$

So the answer for this case would be 12.358 years after.

Step-by-step explanation:

Part a

For this case we know that the decay rate per year is about 17% or 0.17 in fraction and the initial amount is 250 mg for 2009, we can define the model like this:

$A(t) = A_o (1-r)^t$

Where A represent the amount of the substance in mg, r the decay rate = 0.17 and t the number of years after 2009.

Our model for this case would be:

$A(t) = 250 (1-0.17)^t = 250(0.83)^t$

Part b

For this case we have that t= 2023-2009=14 years, so then we can find the amount of substance like this:

$A(t=14) = 250 (0.83)^{14}= 18.408 mg$

Part c

For this case we want to find when the quantity is below 25 mg, so we can do this:

$25 = 250 (0.83)^t$

We can divide both sided by 250 and we got:

$0.1 = 0.83^t$

Now we can apply natural log on both sides and we got:

$ln(0.1) = t ln (0.83)$

And if we solve for t we got:

$t = \frac{ln(0.1)}{ln(0.83)}= 12.358 years$

So the answer for this case would be 12.358 years after.