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
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?
2 answers
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.
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.
1 answer