Question

if joe has 6,000$ and wishes to invest it at 9% for 7 years how much will he have after 7 years, both are compounded continiously

Answer 1

a) if kurt is competing with him and has the same amount of money but wishes to double it in 7 years, what rate should kurt have

Answer 2

Alright so we have a compounded continuoisly problem so

\[\large A = Pe^{rt}\]

P = 6000
r = .09
t = 7

\[\large A = 6000e^{.09 \times 7}\]

Thats for problem 1

For problem 2...we have the same formula...except we dont have 'r' this time
We know he wants double the amount he had to begin with....he wants 12000 dollars so

A = 12000
P = 6000
t = 7

\[\large 12000 = 6000e^{7r}\]

Can you solve these?

Answer 3

so the answer for problem 1 would be 11265.66

Answer 4

Correct!

Answer 5

not 100% sure how to solve part 2, i got up to 12000=6579798.95057^r

Answer 6

do i do the square root

Answer 7

Not quite....for that problem we will use logarithms....

\[\large 12000 = 6000e^{7r}\]

First divide both sides by 6000 what do you have?

Answer 8

2000=e\[^{7r}\]

Answer 9

sorry.... 2000=e^7r

Answer 10

hmmm 12000/6000?

Looks like 2 to me :) right?

Answer 11

ah sorry!! you are correct

Answer 12

No problem...so we have

\[\large 2 = e^{7r}\]

right?

Answer 13

yup

Answer 14

Alright...now, what relationship is there between

\(\large \ln\) and \(\large e\) ?

Answer 15

they are inverse fuctions

Answer 16

Correct...

Meaning...they cancel each other out...so if you took the 'ln' of both sides of this function...you would have

\[\large ln(2) = ln(e^{7r})\]

Now remember a rule

\[\large lna^x = x \times ln(a)\]

So now we have

\[\large ln(2) = 7r \times ln(e)\]

We know since ln and e are inverse functions...so \(\large ln(e)\) just = \(\large 1\)

So we have

\[\large ln(2) = 7r\]

How would you solve for 'r' ?

Answer 17

would you divide ln(2) by 7

Answer 18

^Indeed^

Did everything make sense though?

Answer 19

Yes thank you SOOO MUCH!

Answer 20

No problem :)