Attendence at the firemans ball has been declining. Each year the number attending is 2/3 of the previous years attendence and the cost of food is 3/4 that of the previous year.
In what year will the attendence be less than 1/3 the original attendence?
To solve this problem, we need to analyze the pattern of attendance and identify the year when the attendance will be less than 1/3 of the original attendance.
Let's denote the original attendance as A.
According to the given information, each year the number attending is 2/3 of the previous year's attendance. Therefore, we can represent the attendance for each year as follows:
Year 1: A (Original attendance)
Year 2: (2/3)A
Year 3: (2/3)((2/3)A) = (2/3)^2A
Year 4: (2/3)((2/3)^2A) = (2/3)^3A
...
Year n: (2/3)^(n-1)A
We are looking for the year when the attendance will be less than 1/3 of the original attendance (A).
Setting the attendance less than 1/3A, we have:
(2/3)^(n-1)A < (1/3)A
To simplify the equation, let's divide both sides by A:
(2/3)^(n-1) < 1/3
Taking the logarithm of both sides:
log[(2/3)^(n-1)] < log(1/3)
Using the logarithmic property log(a^b) = b * log(a), this can be written as:
(n-1) * log(2/3) < log(1/3)
Now, we can solve for n:
n - 1 < [log(1/3)] / [log(2/3)]
Using a calculator, we can find the approximated value of [log(1/3)] / [log(2/3)] ≈ 0.6309.
n - 1 < 0.6309
n < 1 + 0.6309
n < 1.6309
Since n represents the number of years, it cannot be fractional. Therefore, rounding up to the nearest whole number, we get:
n ≤ 2
Hence, the attendance will be less than 1/3 of the original attendance in the second year.
Answer: In the second year