Question

the number of cells in a culture is known to double every 5 hours. there are 37 cells at time ( t = 0 ) hours. which of the following is an appropriate model for ( n(t) ), the number of cells in the culture at time ( t ) hours? ( n(t)=37 + 2^{(t / 5)} ) ( n(t)=37 + 2^{(5 t)} ) ( n(t)=37 cdot 2^{(t / 5)} ) ( n(t)=37 cdot 2^{(5 t)} )

Explanation:

Step1: Recall the exponential growth formula

The general formula for exponential growth is \(N(t)=N_0\cdot a^{kt}\), where \(N_0\) is the initial amount, \(a\) is the base (growth factor), \(k\) is a constant related to the time - scale, and \(t\) is time. Here, the initial number of cells \(N_0 = 37\). Since the number of cells doubles, \(a = 2\).

Step2: Determine the value of \(k\)

The cells double every \(5\) hours. When \(t = 5\), \(N(5)=2N_0\). Substitute into \(N(t)=N_0\cdot2^{kt}\), we get \(2N_0=N_0\cdot2^{5k}\). Divide both sides by \(N_0\) (since \(N_0
eq0\)), we have \(2 = 2^{5k}\). Then \(5k = 1\), so \(k=\frac{1}{5}\).

Step3: Write the function

Substitute \(N_0 = 37\) and \(k=\frac{1}{5}\) into the formula \(N(t)=N_0\cdot2^{kt}\), we get \(N(t)=37\cdot2^{\frac{t}{5}}\).

Answer:

C. \(N(t)=37\cdot2^{(t/5)}\)