Question

The function f(x) = 4(2)x represents the growth of a butterfly population every year in a remote swamp. Jan wants to manipulate the formula to an equivalent form that calculates five times a year, not just once a year. Which function is correct for Jan's purpose, and what is the new growth rate?

Answer 1

anyone please help?

Answer 2

@aryandecoolest

Answer 3

You should use ^ to show the x is an exponent, as in
\[ f(x) = 4\cdot 2^x\]

Answer 4

what?

Answer 5

oh, sorry.

Answer 6

4(2)^x, that?

Answer 7

yes. otherwise people might thing you mean 4*2*x (and you would get the wrong answer)

Answer 8

I'm uncertain what you mean by your phrase "calculates five times per year." What does that mean to you? Calculate ... what?..

Answer 9

Ok, do you think you can help me or do you want me type the question?

Answer 10

@aryandecoolest idk, it's a weird written question

Answer 11

I think you use
\[ (1+ \frac{r}{n})^{nx} \]
where r is the original rate and n is now 5

Answer 12

If, instead of typing
f(x)=20(2)x
we were to type
f(x)=20(2)5x....how do you think the behavior of the original function would be modified?

Answer 13

i have read this now 5 times and i am also confused as to what it means "five times a year"

Answer 14

my best guess is that it means \(4\times 2^{\frac{x}{5}}=4\times \sqrt[5]{2}^x\)

Answer 15

now in this case \(x\) would represent one fifth of a year

Answer 16

@phi so it's (1 + r/5)^5(x)?

Answer 17

As much as I'd like to help further, I don't yet see what was done with that '20'.

I do see that if we start with f(x) = 20(2)^x, and insert the multiplier 5, we get
f(x)=20(32)x...

Answer 18

the '20' in the original problem statement represents the original number of butterflies. Note that if x=0, 2^x = 2^0 =1 =f(0)=initial value of f(x).
Were we to insert that factor of 5, the resulting formula would look like
20(2)5x=20(32)x
which would grow a LOT faster....a lot faster than the original formula would, I mean.

Answer 19

yes, I think so. you start with
4(2)^x
write the 2 as (1+1)
4(1+1)^x

now divide the 2nd 1 by 5, and change x to 5x
4(1+1/5)^(5x)

that is my best guess as to what they want

Answer 20

Unfortunately, I share your uncertainty in regard to how to manipulate this last expression so that it resembles one of the four possible answers. ...All I can suggest, at this point, is that you double check to ensure you've copied the problem statement correctly..

Answer 21

that could be right as well, but it is a different function from the original one

Answer 22

(x) = 4(1.15)x; growth rate is 5%

f(x) = 4(1.15)5x; growth rate is 15%

f(x) = 4(2)x; growth rate is 200%

f(x) = 4(2)x, growth rate is 5%
these are my choice and @phi I said D

Answer 23

and @aryandecoolest I did copy this right, it's just flvs wants to use trick words and blah blah stuff that don't make any sense

Answer 24

my guess would give
4(1+1/5)^(5x) which is the same as
4(1.2)^(5x),

however, they came up with 1.15 ?! instead of 1.2

Answer 25

but, thinking about it, they want the same growth rate.
the original says after 1 year we get
4(2)^1 = 4*2 = 8

if we use 4(1.15)^(5x)
after 1 year we get 4(1.15)^5 = 8.045 which is close to 8

in other words, 4(1.15)^(5x) is the correct answer.

Answer 26

did they teach you how to do this type of problem? If so, I would like to see what they say

Answer 27

no they did not

Answer 28

that was my last question, i'm going to see if it's correct and see what they say.

Answer 29

thank you

Answer 30

btw, the answer is choice B
4(1.15)^5x , 15% rate

Answer 31

was it correct @morningskye123