Exponential math question? ?

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

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

:)>

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 

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)