Question

the function f(t)=60000(2)^(t/580) gives the number of bacteria in a population t minutes after an initial observation. how much time, in minutes, does it take for the number of bacteria in the population to double? enter your numerical answer. enter your answer... enter your answer as a number, fraction, or decimal

Explanation:

Step1: Set up the equation

We want to find when the population doubles. Let the initial population be \(f(0)=60000(2)^{\frac{0}{580}} = 60000\). When the population doubles, \(f(t)=120000\). So we set up the equation \(120000 = 60000(2)^{\frac{t}{580}}\).

Step2: Simplify the equation

Divide both sides of the equation by 60000. We get \(\frac{120000}{60000}=(2)^{\frac{t}{580}}\), which simplifies to \(2=(2)^{\frac{t}{580}}\).

Step3: Solve for \(t\)

Since the bases are the same (both are 2), the exponents must be equal. So \(\frac{t}{580}=1\), and then \(t = 580\).

Answer:

580