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R.7 Score: 5.6/7 5/7 answered • Question 5 A population of bacteria is growing according to the equation $P(t) = 900e^{0.07t}$, Estimate when the population will exceed 1208. t= Give your answer accurate to one decimal place. Question Help: Video 1 Video 2 Calculator Submit Question Progress saved Done 0/1 pt 5

          R.7
Score: 5.6/7 5/7 answered
• Question 5
A population of bacteria is growing according to the equation $P(t) = 900e^{0.07t}$, Estimate
when the population will exceed 1208.
t=
Give your answer accurate to one decimal place.
Question Help: Video 1 Video 2
Calculator
Submit Question
Progress saved Done 0/1 pt 5
        
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R.7
Score: 5.6/7 5/7 answered
• Question 5
A population of bacteria is growing according to the equation P(t) = 900e^0.07t, Estimate
when the population will exceed 1208.
t=
Give your answer accurate to one decimal place.
Question Help: Video 1 Video 2
Calculator
Submit Question
Progress saved Done 0/1 pt 5

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Score: (5.6)/(7), (5)/(7) answered Progress saved Question 5 (0)/(1) pt 5 A population of bacteria is growing according to the equation P(t) = 900e^(0.07t). Estimate when the population will exceed 1208. t = Give your answer accurate to one decimal place. Question Help: ◻ Video 1 ◻ Video 2 Calculator R.7 Score: 5.6/7 5/7 answered Progress saved Done Question 5 0/1 pt 5 When will the population exceed 1208? t = Give your answer accurate to one decimal place. Question Help Video 1 Video 2 Calculator Submit Question
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6.1 Practice Problems Question 16 A population of bacteria is growing according to the equation P(t) = 1000e^0.08t. Use a graphing calculator to estimate when the population will exceed 1241. t = Give your answer accurate to one decimal place.

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Transcript

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00:01 So i have a bacteria population given by 1000 e to the 0 .08 t.
00:12 And i want to know when will that population exceed 1241.
00:21 I'm going to start by dividing both sides of my equation by a thousand.
00:26 I'm going to take the natural log of both sides.
00:34 I'm going to bring my exponent down...
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