Math / Algebra

QuestionIf 18 g of a radioactive substance are present initially and 8 yr later only 9.0 g remain, how much of the substance, to the nearest tenth of a gram, will be present after 17 yr?
After 17 yr , there will be $\square$ $g$ of the radioactive substance. (Do not round until the final answer. Then round to the nearest tenth as needed.)

Studdy Solution

STEP 1

What is this asking? How much of the radioactive substance is left after 17 years if it started at 18 grams and was 9 grams after 8 years? Watch out! Don't round off any numbers until the very end!

STEP 2

1. Understand the decay formula
2. Find the decay constant
3. Calculate the remaining substance after 17 years

STEP 3

Alright, let's get into it!
We're dealing with radioactive decay, which follows an exponential decay model.
The formula we use is:
$N(t) = N_0 \cdot e^{-kt}$
where: - $N(t)$ is the amount of substance left after time $t$ , - $N_0$ is the initial amount of the substance, - $k$ is the decay constant, - $t$ is the time in years, - $e$ is the base of the natural logarithm.

STEP 4

We know that initially, $N_0 = 18$ grams.
After 8 years, $N(8) = 9$ grams.
Let's plug these values into the decay formula to find $k$ :
$9 = 18 \cdot e^{-8k}$

STEP 5

To solve for $k$ , first divide both sides by 18:
$\frac{9}{18} = e^{-8k}$
which simplifies to:
$0.5 = e^{-8k}$

STEP 6

Now, take the natural logarithm of both sides to solve for $-8k$ :
$\ln(0.5) = -8k$

STEP 7

Finally, solve for $k$ by dividing both sides by $-8$ :
$k = -\frac{\ln(0.5)}{8}$

STEP 8

Now that we have $k$ , let's find out how much of the substance is left after 17 years.
We'll use the same decay formula:
$N(17) = 18 \cdot e^{-k \cdot 17}$

STEP 9

Substitute the value of $k$ from the previous step:
$N(17) = 18 \cdot e^{-\left(-\frac{\ln(0.5)}{8}\right) \cdot 17}$

STEP 10

Simplify the expression:
$N(17) = 18 \cdot e^{\frac{17 \cdot \ln(0.5)}{8}}$

STEP 11

Calculate the value using a calculator, and remember, do not round until the very end!

STEP 12

After 17 years, there will be approximately $\boxed{4.5}$ grams of the radioactive substance remaining.

Was this helpful?

Studdy Solution

STEP 1

What is this asking? How much of the radioactive substance is left after 17 years if it started at 18 grams and was 9 grams after 8 years? Watch out! Don't round off any numbers until the very end!

STEP 2

1. Understand the decay formula
2. Find the decay constant
3. Calculate the remaining substance after 17 years

STEP 3

STEP 4

We know that initially, $N_0 = 18$ grams.
After 8 years, $N(8) = 9$ grams.
Let's plug these values into the decay formula to find $k$ :
$9 = 18 \cdot e^{-8k}$

STEP 5

To solve for $k$ , first divide both sides by 18:
$\frac{9}{18} = e^{-8k}$
which simplifies to:
$0.5 = e^{-8k}$

STEP 6

Now, take the natural logarithm of both sides to solve for $-8k$ :
$\ln(0.5) = -8k$

STEP 7

Finally, solve for $k$ by dividing both sides by $-8$ :
$k = -\frac{\ln(0.5)}{8}$

STEP 8

Now that we have $k$ , let's find out how much of the substance is left after 17 years.
We'll use the same decay formula:
$N(17) = 18 \cdot e^{-k \cdot 17}$

STEP 9

Substitute the value of $k$ from the previous step:
$N(17) = 18 \cdot e^{-\left(-\frac{\ln(0.5)}{8}\right) \cdot 17}$

STEP 10

Simplify the expression:
$N(17) = 18 \cdot e^{\frac{17 \cdot \ln(0.5)}{8}}$

STEP 11

Calculate the value using a calculator, and remember, do not round until the very end!

STEP 12

After 17 years, there will be approximately $\boxed{4.5}$ grams of the radioactive substance remaining.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

What is this asking? How much of the radioactive substance is left after 17 years if it started at 18 grams and was 9 grams after 8 years? Watch out! Don't round off any numbers until the very end!

STEP 2

1. Understand the decay formula2. Find the decay constant3. Calculate the remaining substance after 17 years

STEP 3

STEP 4

We know that initially, N0=18N_0 = 18N0​=18 grams.After 8 years, N(8)=9N(8) = 9N(8)=9 grams.Let's plug these values into the decay formula to find kkk:9=18⋅e−8k9 = 18 \cdot e^{-8k}9=18⋅e−8k

STEP 5

To solve for kkk, first **divide both sides by 18**:918=e−8k\frac{9}{18} = e^{-8k}189​=e−8kwhich simplifies to:0.5=e−8k0.5 = e^{-8k}0.5=e−8k

STEP 6

Now, **take the natural logarithm** of both sides to solve for −8k-8k−8k:ln⁡(0.5)=−8k\ln(0.5) = -8kln(0.5)=−8k

STEP 7

Finally, **solve for kkk** by dividing both sides by −8-8−8:k=−ln⁡(0.5)8k = -\frac{\ln(0.5)}{8}k=−8ln(0.5)​

STEP 8

Now that we have kkk, let's find out how much of the substance is left after 17 years.We'll use the same decay formula:N(17)=18⋅e−k⋅17N(17) = 18 \cdot e^{-k \cdot 17}N(17)=18⋅e−k⋅17

STEP 9

Substitute the value of kkk from the previous step:N(17)=18⋅e−(−ln⁡(0.5)8)⋅17N(17) = 18 \cdot e^{-\left(-\frac{\ln(0.5)}{8}\right) \cdot 17}N(17)=18⋅e−(−8ln(0.5)​)⋅17

STEP 10

Simplify the expression:N(17)=18⋅e17⋅ln⁡(0.5)8N(17) = 18 \cdot e^{\frac{17 \cdot \ln(0.5)}{8}}N(17)=18⋅e817⋅ln(0.5)​

STEP 11

Calculate the value using a calculator, and remember, **do not round** until the very end!

STEP 12

After 17 years, there will be approximately 4.5\boxed{4.5}4.5​ grams of the radioactive substance remaining.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Understand the decay formula
2. Find the decay constant
3. Calculate the remaining substance after 17 years

We know that initially, $N_0 = 18$ grams.
After 8 years, $N(8) = 9$ grams.
Let's plug these values into the decay formula to find $k$ :
$9 = 18 \cdot e^{-8k}$

To solve for $k$ , first divide both sides by 18:
$\frac{9}{18} = e^{-8k}$
which simplifies to:
$0.5 = e^{-8k}$

Now, take the natural logarithm of both sides to solve for $-8k$ :
$\ln(0.5) = -8k$

Finally, solve for $k$ by dividing both sides by $-8$ :
$k = -\frac{\ln(0.5)}{8}$

Now that we have $k$ , let's find out how much of the substance is left after 17 years.
We'll use the same decay formula:
$N(17) = 18 \cdot e^{-k \cdot 17}$

Substitute the value of $k$ from the previous step:
$N(17) = 18 \cdot e^{-\left(-\frac{\ln(0.5)}{8}\right) \cdot 17}$

Simplify the expression:
$N(17) = 18 \cdot e^{\frac{17 \cdot \ln(0.5)}{8}}$

Calculate the value using a calculator, and remember, do not round until the very end!

After 17 years, there will be approximately $\boxed{4.5}$ grams of the radioactive substance remaining.