BuyFind

Single Variable Calculus: Concepts...

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

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

Answer to Problem 59E

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

Answer to Problem 59E

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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