Exponential math question? ?
2020-05-13 10:54 pm
A population of bacteria triples in size every 5 days. determine how long it will take for the bacteria to grow to 81 times its original size
回答 (4)
2020-05-13 11:17 pm
Method 1:
Original number of bacteria = Nₒ
Final number of bacteria = N
Number of days = d
N/Nₒ = 3^(d/5)
81 = 3^(d/5)
3^(d/5) = 3^4
d/5 = 4
d = 20
Hence, time taken = 20 days
====
Method 2:
Let n be the number of 5-day periods.
3^n = 81
3^n = 3^4
n = 4
Time taken = (5 days) × 4 = 20 days
Original number of bacteria = Nₒ
Final number of bacteria = N
Number of days = d
N/Nₒ = 3^(d/5)
81 = 3^(d/5)
3^(d/5) = 3^4
d/5 = 4
d = 20
Hence, time taken = 20 days
====
Method 2:
Let n be the number of 5-day periods.
3^n = 81
3^n = 3^4
n = 4
Time taken = (5 days) × 4 = 20 days
2020-05-13 11:19 pm
B(t) = B(0) x 3^(t/5)
so, we require when 3^(t/5) = 81
i.e. when 3^(t/5) = 3⁴
Hence, t/5 = 4
so, t = 20 days
:)>
so, we require when 3^(t/5) = 81
i.e. when 3^(t/5) = 3⁴
Hence, t/5 = 4
so, t = 20 days
:)>
2020-05-13 11:18 pm
Let population be p when time t = 0 days
When time t = 5 days population is p*3
When time t = 10 days population is p*3*3 and so on.
When time t = n days population is p*3^(n/5)
What value n is the number of days such that p*3^(n/5) = 81p ?
Since 81 = 3^4 we have 3^(n/5) = 3^4
n/5 = 4, so n = 20
When time t = 20 days population is 81p
When time t = 5 days population is p*3
When time t = 10 days population is p*3*3 and so on.
When time t = n days population is p*3^(n/5)
What value n is the number of days such that p*3^(n/5) = 81p ?
Since 81 = 3^4 we have 3^(n/5) = 3^4
n/5 = 4, so n = 20
When time t = 20 days population is 81p
2020-05-13 10:59 pm
first we have to find how many times the bacteria triples
81/3
=27times
As it triples every 5 days therefore
27 x 5
135 days (Ans)
81/3
=27times
As it triples every 5 days therefore
27 x 5
135 days (Ans)
收錄日期: 2021-04-18 18:32:25
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