A population of 75 foxes in a wildlife preserve quadruples in size every 15 yr. The function

y= 75 * 4^x , where x is the number of 15 yr periods, models the population growth. How many foxes will there be after 45 years?

45 years is 3 periods, so 75*4^3

no answer

What is the function

To find out how many foxes there will be after 45 years, we need to find the value of y when x = 45. We can use the given function y = 75 * 4^x to calculate this.

First, substitute x = 45 into the equation:
y = 75 * 4^45

To simplify this, let's calculate 4^45 separately. Start by calculating the value of 4^1, 4^2, 4^3, and so on until 4^45.

4^1 = 4
4^2 = 4 * 4 = 16
4^3 = 4 * 4 * 4 = 64

Continuing this pattern, we can write 4^45 as:
4^45 = (4^3)^15

Since 4^3 = 64, we have:
4^45 = 64^15

Now, calculate the value of 64^15:
64^1 = 64
64^2 = 64 * 64 = 4,096
64^3 = 64 * 64 * 64 = 262,144

Continuing this pattern, we can write 64^15 as:
64^15 = (64^3)^5

Since 64^3 = 262,144, we have:
64^15 = 262,144^5

Now, calculate the value of 262,144^5:
262,144^1 = 262,144
262,144^2 = 262,144 * 262,144 = 68,719,476,736
262,144^3 = 262,144 * 262,144 * 262,144 = 18,014,398,509,481,216

Continuing this pattern, we can write 262,144^5 as:
262,144^5 = (262,144^3)^2

Since 262,144^3 = 18,014,398,509,481,216, we have:
262,144^5 = 18,014,398,509,481,216^2

Now, calculate the value of 18,014,398,509,481,216^2 using a calculator.

18,014,398,509,481,216^2 = 324,518,553,658,426,726,783,156,020,576

Finally, substitute this value back into the equation y = 75 * 4^x:
y = 75 * 324,518,553,658,426,726,783,156,020,576

Calculating this, we find that there will be approximately 24,339,892,021,386,954,756,236,951,318,000 foxes after 45 years.

4800