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1. A hive of bees contains 20 bees when it is first discovered. After 5 days, there are 29 bees. It is determined that the population of bees increases exponentially. How many bees are will there be after 100 days? Express your answer as an exact value and rounded to the nearest whole number.

2.Can you find the inverse of the function y=4x^2+12+9? Why or why not?
What domain restrictions are necessary to allow the function to have an inverse?
Find the inverse of the domain restricted function.

1. A hive of bees contains 20 bees when it is first discovered. After 5 days, there-example-1
User Rando Shtishi
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1 Answer

22 votes
22 votes

1. ≈ 33759 bees

The parent function for exponential functions is y = a(b^x), where "a" is the initial value and "b" is the growth rate. In this case, "x" is the number of days, and "y" is the number of bees after "x" days. We already know that the initial value is 20, so we can set "a = 20". Next, we know that after 5 days, there are 29 bees, which we can also plug into our equation: "x = 5" and "y = 29". Then, we end up with the equation 29=20(b^5) and solve for the growth rate, "b". To solve for "b", first divide both sides by 20, and then take the 5th root of both sides:

29 = 20(b^5)

29/20 = b^5

^5√(29/20) = ^5√(b^5)

^5√(29/20) = b

Now, we know the equation to this problem is y = 20((^5√(29/20))^x). We can plug in "x = 100" to find the number of bees after 100 days.

y = 20((^5√(29/20))^100)

y ≈ 33759 bees

2. No, you can't find the inverse of the function y=4x^2 + 12x + 9, because, for an inverse to be possible, it must pass the horizontal line test, which this function does not pass. The domain restriction necessary would have to make it so the function passes the horizontal line test, which would mean restricting half of the function from its vertex in this case. The x-value of the vertex (refer to the vertex formula) of this function is -12/8, so we can restrict the domain by only letting the domain be [-12/8, +infinity). Finding the inverse (switch "x" and "y", and solve for "y"):

x = 4y^2 + 12y + 9 x: [-12/8, +infinity)

x - 9 = 4y^2 + 12y

x - 9 = 4y^2 + 12y

(x - 9)/4 = y^2 + 3y

(x - 9)/4 + 2.25 = y^2 + 3y + 2.25

(x - 9)/4 + 2.25 = (y + 1.5)^2

√((x - 9)/4 + 2.25) = y + 1.5

√((x - 9 + 9)/4) = y + 1.5

(√(x) / 2) - 1.5 = y

(-3 + √(x))/2 = y

3a. To solve this problem, we have to find a way to use "log base5 of 3" and "log base5 of 8" to get "log base5 of 2".

First, we can exclude the logbase5 so the problem is easier to look at. We need to find a way to get "2" using "3" and/or "8" (cannot use subtraction/addition in this step, and using "3" and/or "8" is required with multiplication/division). Keep in mind the rules of logs with the same base.

8^1/3 = 2

Now, we put the logbase5 back in:

logbase5 of (8^1/3) = logbase5 of 2

From log rules, we can move the exponent "1/3" down:

1/3 * logbase5 of 8

We know that logbase5 of 8 = 1.2920

1/3 * 1.2920

approximately 0.4307

3b. As previously explained, we need to use "3" and "8" to get to "75/8" while keeping in mind the rules of logs.

First, we can simplify "logbase5 of (75/8)":

logbase5 of (75/8) = (logbase5 of 75) - (logbase5 of 8)

Now, ignore the "logbase5" and use provided values of "3" and "8" to get to 75. Note that we can use the number 5: logbasex of x= 1 [x^1 = x]

(5^2) * 3 = 75

Next, place the "logbase5" back in, and simplify

logbase5 of ((5^2) * 3) - logbase5 of 8

logbase5 of (5^2) + logbase5 of 3 - logbase5 of 8

2logbase5 of 5 + logbase5 of 3 - logbase5 of 8

2(1) + 0.6826 - 1.2920

approximately 1.3906

User Taahira
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