Answer:
(a) [tex]A = A_0 * e^{kt}[/tex]
(b) There will be 1lb left after 14 hours
Step-by-step explanation:
Solving (a): The equation
Since the substance decomposes at a proportional rate, then it follows the following equation
[tex]A(t) = A_0 * e^{kt}[/tex]
Where
[tex]A_0 \to[/tex] Initial Amount
[tex]k \to[/tex] rate
[tex]t \to[/tex] time
[tex]A(t) \to[/tex] Amount at time t
Solving (b):
We have:
[tex]t = 4.6hr[/tex]
[tex]A_0 = 8[/tex]
[tex]A(4.6) = 4[/tex]
First, we calculate k using:
[tex]A(t) = A_0 * e^{kt}[/tex]
This gives:
[tex]A(4.6) = 8 * e^{k*4.6}[/tex]
Substitute: [tex]A(4.6) = 4[/tex]
[tex]4 = 8 * e^{k*4.6}[/tex]
Divide both sides by 4
[tex]0.5 = e^{k*4.6}[/tex]
Take natural logarithm of both sides
[tex]\ln(0.5) = \ln(e^{k*4.6})[/tex]
This gives:
[tex]-0.6931 = k*4.6[/tex]
Solve for k
[tex]k = \frac{-0.6931}{4.6}[/tex]
[tex]k = -0.1507[/tex]
So, we have:
[tex]A(t) = A_0 * e^{kt}[/tex]
[tex]A(t) = 8e^{-0.1507t}[/tex]
To calculate the time when 1 lb will remain, we have:
[tex]A(t) = 1[/tex]
So, the equation becomes
[tex]1= 8e^{-0.1507t}[/tex]
Divide both sides by 8
[tex]0.125= e^{-0.1507t}[/tex]
Take natural logarithm of both sides
[tex]\ln(0.125)= \ln(e^{-0.1507t})[/tex]
[tex]-2.0794= -0.1507t[/tex]
Solve for t
[tex]t = \frac{-2.0794}{-0.1507}[/tex]
[tex]t = 13.7983[/tex]
[tex]t = 14[/tex] --- approximated
