Question

$6500 invested at 6% compounded continuously for 2 years.
A=6500e^.12
then what?

Answer 1

You just evaluate it on a calculator, I suppose..

Answer 2

i cant use a calculator for this class

Answer 3

I don't understand. What are you having trouble with exactly?

Answer 4

That's just sadistic D:

No (normal) human can evaluate e^0.12 without the help of technology (right?)

Answer 5

I agree with terenzreignsz

Answer 6

im supposed to use the natural log and get as close as possible

Answer 7

a woman invests $6500 into an accoun that pays 6% interest per year, compounded continuously what is the amount after 2 years?

Answer 8

I don't know how to do this without a calculator, sorry.

Answer 9

so using A=Pe^rt i have A=6500e^(.06)(2)

Answer 10

what does the A stand for in this formula?

Answer 11

Amount

Answer 12

What country do you live in, if I may ask?

Answer 13

Is E the currency?

Answer 14

so it is in fact my unknown?
USA

Answer 15

e is part of the formula

Answer 16

I think I can help you out

Answer 17

so the next part of the problem is how long will it take for the amount to be $8000
so i would have 8000=6500e^(.06)(t)
then 8000/6500=e^(.06)(t)
then 16/13=e^(.06)(t)
then ln(16/13)=(.06)(t)

Answer 18

And then divide both sides by 0.06, and that's done :D

Answer 19

k so the first part that i was originally asking about is impossible to solve with out a calculator?

Answer 20

if you took ln of both sides

A=6500e^(.12)

\[\Large \ln A = \ln (6500e^{0.12}) = \ln 6500 + 0.12 \ln e\]which you can probably solve for A using a table of logs or something.

But i'd probably seriously consider getting the whole class to agree on telling the professor to F off with this no-calculator bs.

Answer 21

its not just the professor its university wide, math 1050 is a no calculator class, it sucks but its how it is..

Answer 22

You could solve continuously compounded interest with a logarithm table.
I'd help you out some more, but I'm researching this.

Answer 23

thanks wolf.

Answer 24

That seems like a colossal waste of student's time and effort, like forcing students to do long division (other than polynomial) anytime after about 8th grade. Yay for wasted time and energy.

Answer 25

yeah ive been doing this one assignment since 10:00 pm its now 2:15 am
and im still not finished

Answer 26

I looked up a couple of math 1050 courses and a couple don't allow graphing calcs, but allow standard scientific ones. And it seems like the ones that didn't allow calculators had questions that didn't require them (like this one. it does require one).

You're sure it's not just graphing calcs that aren't allowed?

Answer 27

im sure, its no calculators at all
trust me i wish i could use a calculator

Answer 28

maybe my professor only wants me to do part b for this problem cuz i can isolate "t" and get my simplified answer

Answer 29

but idk how he wants me to do part a without a calculator.

Answer 30

kylhea - could you read the private message I just sent you?

Answer 31

thanks wolf

Answer 32

Okay kylhea how much longer will you be online here?

Answer 33

(Just wondering because I need a little time for the research and the calculation.)

Answer 34

Guess I'll start the calculations.

Answer 35

ill be on a while i need to finish this assignment

Answer 36

well actually im gonna need to finish it tomorrow, i gotta be up in 3 hours for work

Answer 37

Okay - check back here tomorrow - or maybe I can send it in a private message.
Have a good night
:-)

Answer 38

ok thanks a ton :) have a great night

Answer 39

you too

Answer 40

Okay I came up with this

6,500 Principal
6.00% Compounded Continuously
2 years

Formula for Converting Annual Rate to Continuously Compounded Rate
Continuously compounded rate = (e^r) -1

where e = 2.7182818285 ('e' is the base of natural logarithms).

Continuously compounded rate = (2.7182818285^.06) -1
Continuously compounded rate = 1.0618365465 -1
Continuously compounded rate = 0.0618365465
(or 6.18365465% as it is reported in newspapers)


Formula for calculating the total amount for an initial amount (principal) invested at a compounded interest rate:
Total = Principal * (1 +r)^years
Total = 6,500 * (1.0618365465)^2
Total = 6,500 * 1.1274968516
Total = 7,328.73

and yes it's just that simple LOL

Answer 41

Continuously compounded rate = (2.7182818285^.06) -1
Continuously compounded rate = 1.0618365465 -1

how'd you do this w/o a calculator?

Answer 42

No I did use a calculator but it can be done with a logarithm table.

Answer 43

Here's one way to calculate continuously compounded interest - but it talks about generating approximate logs and without using without a log table. (Basically you create your own log table.)
http://answers.yahoo.com/question/index?qid=1006051509017

Here's a link to a common logarithm table:
http://www.sosmath.com/tables/logtable/logtable.html
As agentsmith was so anxious to know, here's how to convert a 6% annual rate to a compounded rate:
continuously compounded rate = (e^annual rate) -1

We will have to raise 'e' to the power of .06, then subtract 1.
(2.7182818285^.06) -1
= .06 * log (2.72) (table just has 3 sig figures)
=.06 * 0.4346
=0.026076 now we look up the ANTI-LOG of .026076
(We look at the bold numbers and read to the left)
we see that the answer lies between
1.061 0.02571538 and
1.062 0.02612452
to get the next DECIMAL we look at
the logarithm difference = 0.02612452 -0.02571538 =
0.00040914
Meaning that for every .0001LOG increase,
the ANTI-LOG increases by
0.000040914 (Notice we have decreased this number by 10)

(Yes using Excel here because this is getting WAY too tedious)
Log Anti-Log
1.0610 0.025715380
1.0611 0.025756294
1.0612 0.025797208
1.0613 0.025838122
1.0614 0.025879036
1.0615 0.025919950
1.0616 0.025960864
1.0617 0.026001778
1.0618 0.026042692
1.0619 0.026083606
1.0620 0.026124520

Remember we have to find the ANTI-LOG of 0.026076
Now we see that it lies between 1.0618 and 1.0619
(We should subtract 1 to get .0618 AND .0619)
So, for just that teensy bit of work, we have narrowed down the continuously compounded to 3 significant figures. Wow, what amazing accuracy.

Answer 44

Gee I almost forgot there was a Part B to this problem.
$6,500 invested at 6% compounded continuously for 2 years
how long will it take for the amount to be $8000?
(The 6% interest is (if you remember) is 6.18365465% compounded continuously.)

The compound Interest Formula is:
Total = Principal × (1 + Rate)^years

Solving this for years
Years = {log(total) -log(Principal)} ÷ log(1 + rate)
Years = {log (8,000) -log(6,500)} ÷ log (1.0618365465)
Years = {3.903089987 - 3.8129133566) ÷ 0.0260576689
Years = 0.090176630 ÷ 0.0260576689
Years = 3.460656082

Formulas Used: http://1728.org/compint2.htm
Calculator to check answer: http://1728.org/compint.htm