Question

Trying to answer a compound interest question, Don't need the answer just a little clarification in the steps. 6000=4000(1+r)^5 ?

Answer 1

so are you having to find r...?

Answer 2

Seems as if the person who asked the question didn't stick around for the answer.

Answer 3

Im sorry, yes I am, but I wasn't sure if my steps were correct. Let me post.

Answer 4

Well I have the formula to solve for rate
log(1 + rate) = {log(total) -log(Principal)} ÷ Years
Okay - welcome back

Answer 5

ok

Answer 6

the question was at what interest rate (compounded annually) will a sum of 4000$ grow to 6000$ in 5 years.

Answer 7

so do you have any working...?

Answer 8

so I have to solve for r but, I don't know what im doing,
so I get

6000=4000(1+r)^5
/4000 /4000
1.5= (1+r)^5
1.5^5/2= 1+r ?

Answer 9

well you are correct at

1.5 = (1 + 5)^5

so take the 5th root of both sides of the equation, if you don't know logs

then you have

\[\sqrt[5]{1.5} = 1 + r\]

now solve for r, which will be a decimal, then multiply by 100 to get a percentage

Answer 10

oops should be

1.5 = (1 +r)^5

Answer 11

ohhh.... ok. so if otherwise if I were to use logs which step would I do that. ?

Answer 12

4,000 to 6,000 in 5 years
log(1 + rate) = {log(total) -log(Principal)} ÷ Years
log (1+rate) = [log (6,000) -log(4,000)] / 5
log (1+rate) = [3.7781512504 -3.6020599913] / 5
log (1+rate) = 0.1760912591 / 5
log(1+rate) = 0.0352182518

10^.0352182518 = 1.0844717712 = 1 + rate
rate = .0844717712 or 8.44717712%

Answer 13

i'm sorr@wolf thank you for the help as well, wish I cold give everyone medals.

Answer 14

my suggestion, its easier without logs..

Answer 15

thank you.

Answer 16

Yes it is easier without logs :-)