Problem 68 - Beginning and Intermediate Algebra

Question

The number of bacteria in a culture doubles every day. If a culture begins with 1000 bacteria, how many bacteria are present after 7 days?

Step-by-Step Solution

Answer

There are 128,000 bacteria in the culture after 7 days.

Step 1: Understand the exponential growth formula

In problems involving exponential growth, we can use the formula \(N = N_0 * 2^t\), where \(N\) is the final number of bacteria, \(N_0\) is the initial number of bacteria, and \(t\) is the time (in days, in this case). Since we have all the values except for \(N\), it is a plug and chug problem.

Step 2: Identify the given values

In this case, we are given the initial number of bacteria, \(N_0 = 1000\), and the number of days, \(t = 7\).

Step 3: Substitute the given values into the formula

Substitute the given values for \(N_0\) and \(t\) in the formula: \(N = 1000 * 2^7\)

Step 4: Calculate the result

Now, calculate the value of \(N\): \(N = 1000 * 2^7 = 1000 * 128\) \(N = 128000\) Finally, there are 128,000 bacteria in the culture after 7 days.