Question

Please help me! I'll fan and give a medal!

The function f(x)=16(5)^x represents the growth of a bee population every year in a remote swamp. Jennifer wants to manipulate the formula to an equivalent form that calculates two times a year, not just once a year. Which function is correct for Jennifer's purpose, and what is the new growth rate?

Answer 1

@phi

Answer 2

@Zarkon

Answer 3

@jhonyy9

Answer 4

this one is a bit tricky.

Answer 5

How so?

Answer 6

2 times a year, then make an exponent 2x. In other words multiply the exponent times 2.

Answer 7

I thought that may be it- is there any reason to change the number in the parentheses, or the coeffienct?

Answer 8

I am assuming that every growth, is different, and that 2 growths in one year aren;t same.

Answer 9

I think so- I'm not entirely sure. The problem is kind of vague regarding that...

Answer 10

Or if the 2 growths in the year are same, then make it 10, instead of 5.

Answer 11

Well, that isn't in any of the answer choices, so let's assume that they vary

Answer 12

So, if the 2 growths in a year vary, you will get \(\normalsize\color{black}{ f(x)=16(5)^{2x}}\), and if they don't then you get \(\normalsize\color{black}{ f(x)=16(10)^{x}}\).\(\LARGE\color{white}{ \rm │ }\)

Answer 13

Ok- how would you go about finding the new growth rate?

Answer 14

just solve for x?

Answer 15

Here is how to do it:

\[ 16(5)^x = 16(1+ 4)^x \]
if we double to two per year, we use 2x as the exponent. However, the growth rate (the 4) must change (call it y):
\[ 16(1+ y)^{2x} \]
we want that to give the same number as the original equation.
in other words, after 1 year: 16(5)^1 gives 16*5
we want
\[ 16(1+ y)^{2} = 16 \cdot 5\\ (1+ y)^{2} = 5 \\ 1+y= \sqrt{5} \\ y= \sqrt{5}-1\]

Answer 16

the answer will be
\[ 16(1+\sqrt{5}-1 )^{2x} \\ 16(2.24)^{2x} \]

Answer 17

That makes sense- thank you @phi and @SolomonZelman

Answer 18

with 2 growths the only thing changes that there are 2 growths, but not the overall growth per year ?

Answer 19

I'm not sure what you mean

Answer 20

I mean that he is just basically doing this,
\(\normalsize\color{black}{ f(x)=16(5)^x}\) is equivalent to, \(\LARGE\color{white}{ \rm │ }\)
\(\normalsize\color{black}{ f(x)=16(\sqrt{5}^2)^x}\)\(\LARGE\color{white}{ \rm │ }\)
\(\normalsize\color{black}{ f(x)=16(\sqrt{5})^{2x}}\)\(\LARGE\color{white}{ \rm │ }\)

Answer 21

But the growth per year doesn't change from this

Answer 22

when it should

Answer 23

I am not trying to argue, I am just trying to understand this...

Answer 24

Ok... I don't mean to be dense, but isn't that essentially what you did? xcept he added the radical sign in the parentheses?

Answer 25

Here is a graph of both functions. As you see, they are identical curves.
However, they have different growth rates.

Answer 26

I see, so it is just that you do not change the yearly growth... then yes

Answer 27

So, the growth rate is exactly the same, just written differently?

Answer 28

oi

Answer 29

I thought you change the growth, adding another, additional, growth `×5` so that there are two `×5` growths.....

(misinterpretation )

Answer 30

the growth rate is different as phi showed.

Answer 31

I am just saying what I thought the problem was

Answer 32

Ok! Thank @SolomonZelman

Answer 33

*thanks

Answer 34

Would you mind helping me with a problem an arithmetic sequence?

Answer 35

Anytime... even though I didn't actually help.

Answer 36

Yes, if I know how to do it....

Answer 37

I thought you did :)

Answer 38

What are the explicit equation and domain for an arithmetic sequence with a first term of 6 and a second term of 2?

Answer 39

domain of n, or the terms ?

Answer 40

thinking about it, I made it too complicated. To solve do this
\[ 16(5)^x = 16(5)^{\frac{1}{2} \cdot 2 \cdot x} \\
= 16(5^\frac{1}{2})^{2x} \\
= 16 (\sqrt{5})^{2x}
\]

Answer 41

\(\large\color{black}{ a_1}\) is 6, and \(\large\color{black}{ a_2}\) is 2,. So d= ?

Answer 42

can you find the difference ?

Answer 43

I think so- the answer choices show the explicit equation, then an inequality of n and 1 or 0 (they vary)

Answer 44

4?

Answer 45

Close...
can you post the choices?

Answer 46

You subtract 4, because to get from 6 to 2, you subtract 4. Right? so d=-4.

Answer 47

Sure- give me a moment- I need to type them out and I'm not exactly fast

Answer 48

Okay :)

Answer 49

a sub(n) = 6-2(n-1); all integers where n is greater than or equal to 1

a sub(n) = 6-2(n-1); all integers where n is greater than or equal to 0

a sub(n) = 6-4(n-1); all integers where n is greater than or equal to 0

a sub(n) = 6-4(n-1); all integers where n is greater than or equal to 1

Answer 50

I'm ready for something big here @SolomonZelman ;)

Answer 51

\(\large \color{black}{ a_{n}=6-2(n-1).~~~~\rm ~ all~~integers,~~ n≥1}\) \(\LARGE\color{white}{ \rm │ }\)
\(\large \color{black}{ a_{n}=6-2(n-1).~~~~\rm ~ all~~integers,~~ n≥0}\) \(\LARGE\color{white}{ \rm │ }\)
\(\large \color{black}{ a_{n}=6-4(n-1).~~~~\rm ~ all~~integers,~~ n≥0}\) \(\LARGE\color{white}{ \rm │ }\)
\(\large \color{black}{ a_{n}=6-4(n-1).~~~~\rm ~ all~~integers,~~ n≥1}\) \(\LARGE\color{white}{ \rm │ }\)

Answer 52

yep

Answer 53

exactly

Answer 54

The number in front of the `(n-1)` is the difference. Based on that, exclude A and B
and the domain for n can be 1 or greater, since a term can be
\(\normalsize\color{black}{ 1}\)st, \(\normalsize\color{black}{ 2}\)nd, \(\normalsize\color{black}{3}\)rd... and on.
But, not \(\normalsize\color{black}{ 0}\)th.

Exclude C

Answer 55

So D is my correct answer?

Answer 56

For all the time we spent typing the answers, that took about 10 seconds XD

Answer 57

oi... I have so many more problems. This is a sad day.

Answer 58

But at least we got this prob :) (sorry for a late reply)

Answer 59

It's fine :) Thank you so much for the help! This really is a very generous thing to do

Answer 60

yw