homework 6.1 exponential functions\nscore: 20/140 answered: 2/14\nquestion 3\na population of bacteria is…

homework 6.1 exponential functions\nscore: 20/140 answered: 2/14\nquestion 3\na population of bacteria is growing according to the equation ( p(t)=350 e^{0.23 t} ).\nuse a graphing calculator to estimate when the population will exceed 1491.\n( t= )\ngive your answer accurate to one decimal place.\nquestion help: video

homework 6.1 exponential functions\nscore: 20/140 answered: 2/14\nquestion 3\na population of bacteria is growing according to the equation ( p(t)=350 e^{0.23 t} ).\nuse a graphing calculator to estimate when the population will exceed 1491.\n( t= )\ngive your answer accurate to one decimal place.\nquestion help: video

Answer

Explanation:

Step1: Set up the inequality

Set (P(t)>1491), so (350e^{0.23t}>1491).

Step2: Solve for (e^{0.23t})

Divide both sides by (350): (e^{0.23t}>\frac{1491}{350}\approx4.26).

Step3: Take the natural logarithm of both sides

(\ln(e^{0.23t})>\ln(4.26)). Since (\ln(e^{x}) = x), we have (0.23t>\ln(4.26)).

Step4: Solve for (t)

(t>\frac{\ln(4.26)}{0.23}). Calculate (\ln(4.26)\approx1.45), then (t>\frac{1.45}{0.23}\approx6.3).

Answer:

(t = 6.3)