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?

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

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.