Question

Suppose the number of a kind of bacteria on the nth day (A) can be calculated by the following formula: A=k log n, where k is a constant. If there are 2000 bacteria on the 100th day, when will the amount attain 4000?

Answer 1

@ganeshie8

Answer 2

@ikram002p

Answer 3

soo
A(100)=k log(100)=2000
thus find k :D

Answer 4

Boltzmann ENtity:
S = k log V

Answer 5

k=1000/log 10

then
A(n)=k log n =4000
solve for n

Answer 6

so its 1000

Answer 7

@ikram002p so the answer is 10000th day?

Answer 8

yeah k=1000
next step solve for n
A(n)=1000 log n =4000

Answer 9

so is it 10000

Answer 10

yes :)
n=10^4

Answer 11

A = n^k
2000= (100)^k
\[k \approx 1.6505\]

Answer 12

thanks @ikram002p

Answer 13

np :)

Answer 14

4000 = x^1.6505
x = 152 days

Answer 15

how is that adjax ?

Answer 16

I just converted the expression into exponential form from the log one:
A = n^k
2000 = (100)^k
k = 1.6505

when A = 4000
n = ?
k =1.6505{constant}

4000 = n^1.6505
log 4000 = 1.6506 log n
.....
n = 152 days(rounded-off)

@ikram002p

Answer 17

A=k log n

A=log (n^k)
10^A=n^k

Answer 18

a blunder on my part :k = 2.1505
a little modification and the answer will come currect:
Answer is :
n= 47 days