[Exponential growth & decay]


The half-life of radioactive strontium-90 is approximately 29 years. In 1965, radioactive strontium-90 was released into the atmosphere during testing of nuclear weapons, and was absorbed into people's bones. How many years does it take until only 14 percent of the original amount absorbed remains?

time= ____ years

Many math-teachers use base e for most exponential problems of this type, but since "half-life" is involved, let's use 1/2 as the base

amount = a(1/2)^(t/k) or
amount = a(2)^(-t/k) where a is the initial amount, t is the time in years, and k will be 29

so we have to solve
.14 = 1(2)^(-t/29(
take the ln of both sides
ln .14 = -t/29(ln 2), hope you know your log rules
-t/29 = ln .14/ln 2
-t/29 = -2.8365
t = 82.26 years.

check: after 29 years, we have 50% left
after 58 years we have 25% left
after 87 years we would have 12.5% left.

our answer appears reasonable.

Well, isn't this a bone-tickling question! Let's put on our mathematical lab coats and solve it.

Since the half-life of radioactive strontium-90 is 29 years, we can use the formula for exponential decay to figure out the time it takes for only 14 percent of the original amount to remain.

Now, if 50 percent (half) of the strontium-90 decays in 29 years, then 14 percent should decay even faster. So, we need to figure out how many half-lives it takes to reach that delightful 14 percent mark.

By dividing 100 percent by 14 percent, we find that we have about 7.14 half-lives.

Since each half-life is 29 years, we multiply 29 by 7.14 to find out what hilariously obscure number of years it will take for only 14 percent of the original amount absorbed remains.

So, the answer is approximately 206.06 years. But don't worry, with all the clowning around I'll be doing, time will fly by like a laughing cannonball!

To determine the time it takes until only 14 percent of the original amount of radioactive strontium-90 remains, we can use the formula for exponential decay:

Nt = N0 * (1/2)^(t/h)

Where:
Nt is the amount remaining after time t
N0 is the initial amount
t is the time elapsed
h is the half-life

In this case, we want to find the time when only 14 percent (0.14) of the original amount remains. The initial amount is 100 percent (1), so we have:

0.14 = 1 * (1/2)^(t/29)

To solve for t, we can take the logarithm (base 2) of both sides:

log2(0.14) = log2((1/2)^(t/29))

Using the logarithmic property log2(x^y) = y * log2(x), we get:

log2(0.14) = (t/29) * log2(1/2)

Simplifying further:

log2(0.14) = (-t/29) * log2(2)

Since log2(2) = 1, we can simplify to:

log2(0.14) = -t/29

To find t, we multiply both sides by -29:

-29 * log2(0.14) = t

Using a calculator, we can find the value of -29 * log2(0.14) is approximately -48.681.

Therefore, it takes approximately 48.681 years for only 14 percent of the original amount absorbed to remain.

To determine the number of years it takes until only 14 percent of the original amount absorbed remains, we can use the concept of exponential decay and the half-life of radioactive strontium-90.

The half-life of strontium-90 is approximately 29 years, which means that after every 29 years, half of the amount of strontium-90 will decay. Since we want to know when only 14 percent of the original amount remains, it means we are interested in finding the number of half-lives that have passed until that point.

To calculate the number of half-lives, we can use the following formula:

Number of half-lives = (ln(final amount / initial amount)) / (ln(0.5))

In this case, the final amount is 14 percent of the initial amount, which can be written as 0.14. The initial amount is 100 percent or 1. Plugging these values into the formula:

Number of half-lives = (ln(0.14) / ln(0.5))

Using a scientific calculator or logarithm tables, we can find the natural logarithm (ln) values and evaluate this expression.

After calculating the number of half-lives, we can multiply it by the half-life duration of 29 years to obtain the final answer in years.

time = (number of half-lives) * (half-life duration)

I don't have the capability to use logarithms or evaluate expressions, but you can try using a scientific calculator or logarithm tables to find the number of half-lives and then calculate the time it takes until only 14 percent of the original amount absorbed remains.