Question

Every day, the number of fish in a pond doubles. if it takes 24 days to fill the entire pond with fish, how long does it take to fill half the pond?

Answer 1

Say initial population occupies 1/n of the pond.
 Given: (1/n)∗(2^24)=1
 Question: if (1/n)∗(2^x)=1/2 then x=?
 (1/n)∗(2^x)=1/2 --> (1/n)∗(2^x)∗(2)=1 --> (1/n)∗(2^(x+1))=1
 Since we know that
 (1/n)∗(2^24)=1
  then 
 2^(x+1)=2^24  --> x+1=24  --> x=23
 take 23 days to fill half the pond 
 

Answer 2

Assume its size is 1 unit at birth.
On day 1 its size is 2 units in size.
On day 2 its size is 4 units in size.

Formula is 1 * 2^n where n is the number of days of its life. 

On day 1, it is 1*2^1 = 2 units in size.
On day 2, it is 1*2^2 = 4 units in size. 

On day 24, it is 1^2^24 = 16777216 units in size. 

The pond is covered when the lily pad is 16777216 units in size. 

The pond is half covered when the lily pad is 16777216/2 = 8388608 units in size. 

The number of days it takes for the lily pad to become 8388608 units in size is given by the formula: 

8388608 = 1 * 2^x 

you need to solve for x. 

Take the log of both sides of this equation to get: 

log(8388608) = log(1*2^x) 

Since 1*2^x is the same as 2^x, this equation becomes: 

log(8388608) = log(2^x) 

Since log(2^x) = x*log(2), this equation becomes: 

log(8388608) = x*log(2) 

Divide both sides of this equation by log(2) to get: 

log(8388608)/log(2) = x 

divide the two, you’ll get 23. The pond was half covered with the lily pad on the 23rd day.