if someone offers to pay you 1 penny on the first day, 2 pennies on the second day, 4 pennies on the third day, 8 pennies on the fourth day, 16 pennies on the fifth day ad so on. What is the first day on which your pay would be at least $100.
The pay on the nth day is 2^(n-1)
So,we want
2^(n-1) = 10000
n-1 = log10000/log2
n = 14.28
So, on the 15th day the pay is 2^14 = $163.84
To find the first day on which your pay would be at least $100, we can use a formula to calculate the pay for each day. The amount of pay doubles each day, starting with 1 penny on day 1.
The formula for the pay on a particular day can be expressed as 2^(n-1), where n is the day number.
Now, let's find the day on which the pay is at least $100. We need to solve the following equation:
2^(n-1) >= 100
To get the value of n, we can take the logarithm base 2 of both sides of the equation:
log2(2^(n-1)) >= log2(100)
Simplifying the equation further:
n-1 >= log2(100)
Now, let's solve for n:
n >= log2(100) + 1
Using a calculator, we find:
n >= 6.643
Since the day number must be a whole number, we round up to the nearest whole number to get the minimum day number:
n = 7
Therefore, the first day on which your pay would be at least $100 is day 7.