10)Suppose the amount of Carbon-14 remaining in a sample of 100 milligrams after xyears can be described by A(x) = 100 e-0.01294xa) How many milligrams of Carbon-14 are initially present: 1b) How many years will it take for the amount of Carbon-14 to decay to 3milligrams?

Answered: 10)Suppose the amount of Carbon-14…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section: Chapter Questions

Problem 5T

See similar textbooks

Related questions

Question

10)Suppose the amount of Carbon-14 remaining in a sample of 100 milligrams after x
years can be described by A(x) = 100 e-0.01294x
a) How many milligrams of Carbon-14 are initially present: 1
b) How many years will it take for the amount of Carbon-14 to decay to 3
milligrams?

Expert Solution

Step 1

Given the amount of Carbon-14 remaining in a sample of 100 milligrams after $x$ years is given by:

$A (x) = 100 e^{- 0.01294 x}$

To find:

a) Amount of Carbon-14 initially present.

b) Time required to decay to 3 milligrams.

Solution:

Initial amount means amount at $x = 0$ .

Putting $x = 0$ in given equation of the amount.

$\begin{array}{rcl} A (0) & = & 100 e^{- 0.01294 \cdot 0} \\ = & 100 e^{0} \\ = & 100 \cdot 1 \\ = & 100 \end{array}$

Therefore, amount of carbon present initially is100 milligrams.

Step 2

Now, when amount of Carbon -14 decay to 3 milligrams then it means $A (x) = 3$ .

Therefore,

$3 = 100 e^{- 0.01294 x}$

Taking logarithm on both sides, we get

$\begin{array}{rcl} 3 & = & 100 e^{- 0.01294 x} \\ \ln (3) & = & \ln (100 e^{- 0.01294 x}) \\ \ln 3 & = & \ln 100 + \ln e^{- 0.01294 x} \\ \ln 3 & = & \ln 10^{2} + (- 0.01294 x) \ln e [Using \ln a^{m} = m \ln a] \\ 1.098 & = & 2 \ln (10) - 0.01294 x \cdot 1 \\ 1.098 & = & 2 \cdot 2.302 - 0.01294 x \\ 1.098 & = & 4.604 - 0.01294 x \\ 0.01294 x & = & 3.506 \\ x & = & \frac{3.506}{0.01294} \\ x & = & 270.94 \end{array}$