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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 5: Q 9 (page 336)

Radioactive Decay
The half-life of radium is 1690 years.
If 10 grams is present now, how much will be present in 50 years?

Short Answer

Expert verified

The amount of radium present in 50 years is 9.797 grams.

Step by step solution

01

Step 1. Given information

The half-life of radium is 1690 years.

If 10 grams is present now, how much will be present in 50 years

02

Step 2. Radium present in 50 years

The amount A of radioactive material present at time t is expressed by the function $N (t) = N_{0} e^{k t}$ where $k < 0$ represents the rate of decay.

The half-life of radium is 1690 years.

Substitute $N_{0} = 10, N (t) = \frac{10}{2} = 5, t = 1690$ in the function $N (t) = N_{0} e^{k t}$

role="math" localid="1647290301897" $N (t) = N_{0} e^{k t} 5 = 10 e^{k 脳 1690} \frac{1}{2} = e^{k 脳 1690} \ln (0.5) = \ln (e^{k 脳 1690}) \ln (0.5) = k 脳 1690 k = \frac{\ln (0.5)}{1690} k = - 0.00041$

Now radium present in 50 years can be calculated by,

substituting role="math" localid="1647290364834" $N_{0} = 10, k = - 0.00041, t = 50$ in the function $N (t) = N_{0} e^{k t}$

$N (t) = N_{0} e^{k t} N (50) = 10 e^{- 0.00041 脳 50} N (50) = 9.797$

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Most popular questions from this chapter

Begin with the graph of $y = e^{x}$ and use transformation to graph the function. Determine the domain, range and horizontal asymptote of the function.

localid="1646160286433" $f (x) = 5 - e^{- x}$

solve each equation. Verify your results using a graphing utility.

$4^{x^{2}} = 2^{x}$

solve each equation. Verify your results using a graphing utility.

$4^{x} 2^{x^{2}} = 16^{2}$

True or False $f (g (x)) = f (x) . g (x)$ .

solve equation. Verify your results using a graphing utility.

$8^{- x + 14} = 16^{x}$

See all solutions

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