Question

The number of bacteria in a petri dish doubles every 4 hours. If there are initially 200 bacteria.

a) How many bacteria will there be after 12 hours?

b) How many bacteria will there be after 2 days?

Answer 1

y = A bᵗ

The starting amount of bacteria is 100, so A = 100

To find b, plug in this equation, t = 4 = and y = 200 (since the population doubles in 4 hours)

200 = 100 ∙ b⁴

Divide both sides by 2

200 / 100 = b⁴

b⁴ = 2

b = ∜2

y = A bᵗ

y = 100 ∙ (∜2 )ᵗ

a)

After 12 h:

y = 100 ∙ (∜2 )¹²

Since (∜2 )¹² = 2³

100 ∙ 2³ = 100 ∙ 8 = 800

b)

2 days = 48 h

y = 100 ∙ (∜2 )⁴⁸

Since (∜2 )⁴⁸ = 2¹²

y = 100 ∙ 2¹² = 100 ∙ 4096 = 409 600

Answer 2

The previous post is wrong because I took the wrong starting value of the bacteria.

The solution should be written as follows:

y = A bᵗ

The starting amount of bacteria is 200, so A = 200

To find b, plug in this equation, t = 4 = and y = 400 (since the population doubles in 4 hours)

y = A bᵗ

y = 100 bᵗ

400 = 100 ∙ b⁴

Divide both sides by 4

400 / 100 = b⁴

b⁴ = 4

b = ∜4

b = √2

y = A bᵗ

y = 100 ∙ √2ᵗ

a)

After 12 h:

y = 100 ∙ ( √2 )¹²

Since ( √2 )¹² = 2⁶

100 ∙ 2⁶ = 100 ∙ 64 = 6 400

b)

2 days = 48 h

y = 100 ∙ ( √2 )⁴⁸

Since ( √2 )⁴⁸ = 2²⁴

y = 100 ∙ 2²⁴ = 100 ∙ 16 777 216 = 1 677 721 600

Ignore my first post.