# The inverse of the function.

### 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 1.6, Problem 59E

(a)

To determine

## To find: The inverse of the function.

Expert Solution

The inverse of the function is f1(t)=4.328ln(t100)

### Explanation of Solution

Given information:

Initial population of the bacteria is 100 and the population is double in every three hours.

The inverse of the function is calculated as,

f(t)=100(2t/3)y=1002t/32t/3=y100t3ln2=ln(y100)

Simplify further,

t=4.328ln(y100)

For the inverse function, replace t and y ,

y=4.328ln(t100)f1(t)=4.328ln(t100)

Therefore, the inverse of the functionis f1(t)=4.328ln(t100) .

(b)

To determine

### To find: The number of hours in which population of the bacteria reach 50,000 .

Expert Solution

The number of hours in which population of the bacteria reach 50,000 is 26.89h .

### Explanation of Solution

Given information:

Initial population of the bacteria is 100 and the population is double in every three hours.

The number of hours in which population of the bacteria reach 50,000 is calculated as,

t=4.328ln(y100)=4.328ln(50000100)=26.89h

Therefore, the number of hours in which population of the bacteria reach 50,000 is 26.89h .

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