Question

In a lab experiment, 80 bacteria are placed in a petri dish. The conditions are such that the number of bacteria is able to double every 5 hours. How long would it be, to the nearest tenth of an hour, until there are 136 bacteria present?

Answer 1

It would take approximately 8.2 hours until there are 136 bacteria present.

Explanation:

The exponential growth of bacteria can be modeled by the formula
`\(N(t) = N_0 \cdot 2^((t/h))\)`, where
`\(N(t)\)` is the number of bacteria at time
`\(t\), \(N_0\)`is the initial number of bacteria,
`\(t\)` is the time elapsed, and
`\(h\)` is the doubling time.

In this case,
`\(N_0 = 80\), \(N(t) = 136\), and \(h = 5\)` hours. The formula becomes
`\(136 = 80 \cdot 2^((t/5))\).`

To find the time
`(\(t\)),` we can rearrange the equation:
`\((136)/(80) = 2^((t/5))\).`

Solving for
`\(t\),`we take the logarithm of both sides. Using the logarithm base 2, we get
`\( (t)/(5) = \log_2\left((136)/(80)\right) \).`

Multiplying both sides by 5 gives
`\(t = 5 \cdot \log_2\left((136)/(80)\right)\).`

Calculating this expression, we find
`\(t \approx 8.2\)` hours.

Therefore, it would take approximately 8.2 hours until there are 136 bacteria present.

Answer 2

Explanation:

We can use the formula for exponential growth to solve this problem:

N = N0 * 2^(t/T)

where N is the final number of bacteria, N0 is the initial number of bacteria, t is the time elapsed, and T is the doubling time.

We know that N0 = 80, and T = 5 hours. We want to find t when N = 136.

136 = 80 * 2^(t/5)

Dividing both sides by 80, we get:

1.7 = 2^(t/5)

Taking the logarithm of both sides (base 2), we get:

log2(1.7) = t/5

Solving for t, we get:

t = 5 * log2(1.7) ≈ 3.4 hours

Therefore, it would take approximately 3.4 hours, to the nearest tenth of an hour, until there are 136 bacteria present.