Question

If $535 is invested at an interest rate of 6% per year and is compounded continuously, how much will the investment be worth in 10 years? So I plugged that in just like someone showed me and I got zero??

Answer 1

remember
Pe^rt

Answer 2

whenever they say continous compounding

Answer 3

\[A=Pe^{rt}\]

Answer 4

A= 535 e^(.06*10)
now just evaluate this

Answer 5

Okay so wait what about e?

Answer 6

most calculator has 'e' button

Answer 7

Okay, so once I multiply .06 *10 what do I do next? divide or multiply? Oh okay, so I look for e. alright that makes sense.

Answer 8

I thought I had tos olve for e or something.

Answer 9

A= 535 e^(.6)

Answer 10

http://www.wolframalpha.com/input/?i=535+e^%28.6%29

Answer 11

Wow, thanks :). So for: A bacteria culture begins with 4 bacteria which double in size every hour. How many bacteria exist in the culture after 8 hours? I'd do A=4 e ^(.8)

Answer 12

\[A=A_0 e^{rt}\]
This is the formula
\[A_0\] is initial state
-----------------------------------------------
\[r\]
is rate
---------------------------------------------
\[t\]
is time

Answer 13

we have to find r first

Answer 14

2? ^2?

Answer 15

every 60 minutes?

Answer 16

A bacteria culture begins with 4 bacteria which double in size every hour

\[A=A_0e^{rt}\]
\[8=4e^{r1}\]
\[2=e^{r1}\]
\[ln(2)=ln(e^r)\]
\[ln(2)=r * ln (e)\]
\[ln(2)=r \]

Answer 17

So are they two different equations/

Answer 18

actually , they are same we just use different letters

Answer 19

So so i plug anything into ln(2)=r

Answer 20

yes,

Answer 21

Or I can just plug it into a=ae^rt

Answer 22

\[A=A_0 e^{ln(2)t}\]

Answer 23

Is the answer 1024?

Answer 24

That's what I got.

Answer 25

\[A=4e^{ln(2)8}\]
Yes 1024

GJ

Answer 26

Thank you soo much :)