Question

Substance A decomposes at a rate proportional to the amount of A present. a) Write an equation that gives the amount A left of an initial amount A0 after time t. b) It is found that 8 lb of A will reduce to 4 lb in 4.6 hr After how long will there be only 1 lb left?
a) Choose the equation that gives A in terms of A0, t, and k, where k > 0.
b) There will be 1 lb left after 14 hr (Do not round until the final answer. Then round to the nearest whole number as needed.)

Answer 1

Answer:

(a) $A = A_0 * e^{kt}$

(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

$A(t) = A_0 * e^{kt}$

Where

$A_0 \to$ Initial Amount

$k \to$ rate

$t \to$ time

$A(t) \to$ Amount at time t

Solving (b):

We have:

$t = 4.6hr$

$A_0 = 8$

$A(4.6) = 4$

First, we calculate k using:

$A(t) = A_0 * e^{kt}$

This gives:

$A(4.6) = 8 * e^{k*4.6}$

Substitute: $A(4.6) = 4$

$4 = 8 * e^{k*4.6}$

Divide both sides by 4

$0.5 = e^{k*4.6}$

Take natural logarithm of both sides

$\ln(0.5) = \ln(e^{k*4.6})$

This gives:

$-0.6931 = k*4.6$

Solve for k

$k = \frac{-0.6931}{4.6}$

$k = -0.1507$

So, we have:

$A(t) = A_0 * e^{kt}$

$A(t) = 8e^{-0.1507t}$

To calculate the time when 1 lb will remain, we have:

$A(t) = 1$

So, the equation becomes

$1= 8e^{-0.1507t}$

Divide both sides by 8

$0.125= e^{-0.1507t}$

Take natural logarithm of both sides

$\ln(0.125)= \ln(e^{-0.1507t})$

$-2.0794= -0.1507t$

Solve for t

$t = \frac{-2.0794}{-0.1507}$

$t = 13.7983$

$t = 14$ --- approximated