Question

On his way home from the laboratory, Louie realized that he left a test tube containing 2560 yeast cells in the lab. Every two minutes, the number of cells in the test tube increases by 50 percent. If the number of cells reaches 98,415, the test tube will explode! Naturally, he turned around and rushed back to the lab. It took Louie t minutes to return to the lab, and he found the test tube intact. Write an inequality in terms of t that models the situation.

Answer 1

Answer: 2560$e^{0.2027.t}$ < 98415

Step-by-step explanation:

The number of cells increase in an exponential growth, which is, in general:

A(t) = A₀.$e^{kt}$

where;

A is the growth at a time "t"

A₀ is the initial amount of cells

k is rate of growth

t is time in minutes

To write an equation for the conditions described above, we have to find the rate k, knowing that at every 2 minutes, the number of cells increases by 50%, i.e., A₀*1.5:

A(2) = 2560$e^{2k}$

2560*1.5 = 2560$e^{2k}$

$e^{2k}$ = 1.5

ln($e^{2k}$) = ln(1.5)

k = 0.2027

With the initial value, the rate and knowing that the number of cells has to be less than 98415:

2560$e^{0.2027.t}$ < 98415

The inequality in terms of t is 2560$e^{0.2027.t}$ < 98415

Answer 2

Answer:

the answer is 2560*(1.5)^t/2<98415

Step-by-step explanation: