# The quantity of mass that remains after 16days.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 1, Problem 27RE

(a)

To determine

## To find: The quantity of mass that remains after 16days.

Expert Solution

Solution:

The quantity of mass that remains after 16days is 116g.

### Explanation of Solution

Given:

The initial mass of the sample = 1g.

The half-life of Palladium-100 (100Pd) = 4 days.

Calculation:

Let the mass be m.

The initial mass = m and the mass of Palladium-100 in 4days = 12m.

The mass after 8days=12(12m)=(12)2m=14m.

The mass after 12days= 12(12)2m=(12)3m=18m.

The mass after 16days=12(12)3m=(12)4m=116m.

Since the initial mass m = 1g, the quantity of mass after 16 days is, 116(1g).

Thus, the mass that remains after 16days is 116g.

(b)

To determine

### To find: The mass that remains after t days.

Expert Solution

Solution:

The mass that remains after t days is 2t4.

### Explanation of Solution

Given:

The initial mass of the sample=1g.

The half-life of Palladium-100 (100Pd) = 4days.

Calculation:

From part (a), the mass after 16 days is, (12)164m=116g.

The general form of the mass after n days is given by, (12)n4

Therefore, the mass remains after t days is (12)t4.

(c)

To determine

### To find: The inverse of (12)t4 and explanation of its meaning.

Expert Solution

Solution:

The inverse of the function is f1(m)=4log2(m) and the meaning of the inverse function in which the time is passed when m gram is left.

### Explanation of Solution

Laws of logarithms:

If x and y are positive numbers, then

1.logb(xr)=rlogbx2.logb(b)=1

Calculation:

Let m(t)=2t4 be the mass that remains after t days.

Then its inverse function is, t=f1(m).

Here, m=2t4.

Taking log2 on both sides and simplify as follows.

log2(m)=log2(2t4)=t4log2(2)=t4(1)[by laws of logarithms]=t4

Simplify further as,

log2(m)=t44log2(m)=t t=4log2(m)

Substitute t value in the equation (2)

Thus, f1(m)=4log2(m).

Therefore, the inverse of the function is f1(m)=4log2(m) it represents the time when is passed when m grams left.

(d)

To determine

### To find: When the mass is reduced to 0.01g.

Expert Solution

Solution:

The mass is reduced to 0.01g approximately in 26.6 days.

### Explanation of Solution

Calculation:

From part (a), m=2t4.

Substitute m=0.01 in m=2t4,

0.01=2t4

Taking log on both sides.

log(0.01)=log(2t4)log(102)=t4log(2)2log(10)=t4(log2)2=t4(log2)

On further simplification,

2×4log2=tt=8log2t26.6

Therefore, the mass is reduced to 0.01g approximately in 26.6 days.

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!