Question

The number of bees in a hive is 1000 on June 1 and doubles every month. This can be expressed as N=1000 X 2^t, where N represents the number of bees and t represents time, in months.
a) Find the number of bees after 2,3,4, and 5 months
b) What does t=0 represent in this situation?
c)Is it possible for t to be -1? What does this mean?
d) When were there 125 bees? Explain.

Answer 1

N=1000 x 2^t
x is not another variable, it's multiplication?

Answer 2

yes it's multiplication

Answer 3

have you tried doing part a at least?

Answer 4

I just have to sub 2,3,4 and 5 for t right?

Answer 5

Yes

Answer 6

To get N which represents the number of bees

Answer 7

1000x2^2=4000 in 2months
1000x2^3=8000 in 3 months
1000x2^4=16,000 in 4 months
10000x2^5=32,000 in 5 months

Answer 8

I don't really understand part b

Answer 9

@Zale101 ^

Answer 10

b) What does t=0 represent in this situation?
What happens to the function when t=0?

Answer 11

1000 x 2^0=1000
the number of bees stays the same when t is 0? idk

Answer 12

In your question, it says "The number of bees in a hive is 1000 on June 1 "
So, when t=0, the number of bees is 1000. That is the same number of bees starting on June 1.

Answer 13

ohh i see

Answer 14

for part c...
1000 X 2^-1=500

Answer 15

Are those the number of bees before June 1st?

Answer 16

Yes.

Answer 17

For part D, all you have to do is sub 125 to n and solve for t using logs.

Answer 18

okay so 125=1000 x2^t
but what are logs?

Answer 19

These are logs https://www.mathsisfun.com/algebra/logarithms.html

Answer 20

\( log_{10}(\frac{125}{1000})=log_{10} 2^t\)

Answer 21

I've never learned about logs before
based on the example would be it be 5x5x5=125
i'm just guessing

Answer 22

125/1000 is reduced to 1/8
\(log_{10}(\frac{1}{8}=log_{10}2^t\)

Answer 23

well, try to do further study on logarithms because you might get tested on it.

Answer 24

@Zale101 so would my answer be log10(1/8=log10 2t

Answer 25

or..

Answer 26

something else

Answer 27

\((log_{10}(1/8))= (log_{10}(2^t))\)
\({log_{10}\frac{1}{8}}=t~log_{10}2\)
\(\LARGE \frac{(log_{10}(\frac{1}{8})}{(log_{10}(2))}=\large t\)

Answer 28

what would that equal to?

Answer 29

somehow i got 1/4

Answer 30

http://www.chilimath.com/algebra/advanced/log/images/466x428xrules,P20of,P20exponents.gif.pagespeed.ic.1z7-Ekl5L5.png Here are some log rules for you to know and learn of how i solved it.

Answer 31

The answer is not 1/4

Answer 32

http://www.wolframalpha.com/input/?i=%28log_10%281%2F8%29%29%2F%28log_10%282%29%29%3Dt
You can just plug it to the calculator or wolfram.

Answer 33

@Zale101 alright so the final answer is -3?