Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Radioactive Carbon 14 decays by about $11 \%$ every 1000 years. How much would remain in a 50 lb sample at the end of (a) 5000 years (b) 20,000 years?

(a) $27.9203 \mathrm{lb}$(b) $4.8615 \mathrm{lb}$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 2

Exponential Functions

Campbell University

McMaster University

Baylor University

Lectures

02:11

A 12.0-g sample of carbon …

02:22

A $12.0 \mathrm{~g}$ sampl…

02:14

Radium decays by about $35…

02:05

The isotope carbon-14, &#x…

03:14

43.23. A $12.0-\mathrm{g}$…

06:42

03:08

01:39

A radioactive substance ha…

02:19

The quantity $Q$ of radioa…

02:47

Carbon-14 Decay The rate a…

01:03

Half-life of radioactive $…

02:30

A radioactive element deca…

07:27

$\bullet \mathrm{A} 12.0$ …

02:08

Radioactive carbon- 14 is …

so if we want to figure out how long it will take or how much will have left after each of these time periods, One thing we need to do since our rate isn't in per year but it's in per thousands of years will need to slightly modify our equation. So normally we have P F T is equal to peanut, one plus are raised deep. So our T is normally in years. But now, since our decay or our rate is in thousands of years, we're going to want t to be equal to 1000 years. So that means for a If we want 5000 years later, we would divide that by 1000. So we just have t 0 to 5 and then likewise here, divide that by 1000 and we get 20. So these are the numbers we would be plugging into our equation up here. So first, our initial amount is just gonna be 50 and there will be one, and then our our is going to be Well, it's a DK, so our should be negative, and then we convert 11% to a decimal, which would just be point 11 negative 0.11 Race city. So we get 50 raised to the 0.89 T. Now we just need to plug in five and 20 into this equation. And doing that should give us how much we have left in each case. Um, so 0.89 4.8959 rays of the five multiply that by 50. So we end up with something around £27.92 and then in the second case, we do p of 20 which is going to be 50 times 0.89 Race the 20. So 0.89 raised to the 20 times 50 and that gives us something around 4.86 pounds. So this is how much we would have in the case of after 5000 and 20,000 years.

View More Answers From This Book

Find Another Textbook

Numerade Educator

01:58

Solve for $x.$$$\log x+\log (x+1)=\log 12$$

01:02

Find $x$.$$\log _{25} x=1 / 2$$

Write the given expression in exponential format. $$\log 100=2$$

01:12

Write the given expression as a sum of logarithms.$$\log 3 x$$

02:01

Solve for $x$ in.$$\frac{3^{2 x-1} 9^{1+3 x}}{81^{2+3 x}}=27^{5-4 x}$$

00:52

Write the given expression in logarithmic format.$$6^{0}=1$$

01:20

Given the functions you found in Example $60,$ suppose you are later given a…

01:01

Determine $f^{\prime}(x)$.$$f(x)=\ln x^{n}$$

08:53

(a) Find the equation of the tangent line to the curve $x^{3}+3 y^{2}-12 x-1…

02:26

Sketch on the same graph $y=f(x)=2^{x}$ and $y=g(x)=\frac{1}{2^{x}}=2^{-x} .…