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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?

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  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

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    Bosnian

  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.

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