Question

How to solve compound interest problems?
39,145(1+0.03/12)^12t = 10,000
I know that t will be negative, but no idea how to solve.

Answer 1

here amount is less than principal, which is impossible.

Answer 2

I'm trying to figure out how many months it will be until the principal is paid down to $10,000.00

Answer 3

can you copy the full statement?

Answer 4

You are going to purchase a new car, but being a responsible consumer means doing a little bit of research first.

First, you find the vehicle you are purchasing and its price.
Vehicle: Chevy Volt
Price: $39,145

Current interest rate: 3%

Using the function A(t)=P(1+ r/n)^nt, create the function that represents your new car loan that is compounded monthly. The principle will be the price of the vehicle you selected, not how much you are putting down.
Being a smart financial planner, you want to figure out how many months it will be until your principal is paid down to $10,000.00. Solve for t and show all of your work. Note that t will be negative because the number of months will decrease the principal.
Lastly, you decide to keep track of your loan four times a month instead of monthly. Solve for the adjusted interest rate. Remember to use the formula A(t)=P[(1+ r/n)^1/c]^cnt where c = 4. When solving for the adjusted interest rate, be sure to set it equal to 1+ r/n.

Answer 5

@OnePieceFTW
Have never seen it done this way, but it works!

Answer 6

Is this way harder?

Answer 7

The same answer if you do it the conventional way, but the answer is negative because we're talking about the past.

Answer 8

\[\log 39145 = 12t \log 1.0025\] is this anywhere close on how I should solve this?

Answer 9

Good, so can you now solve for t?

Answer 10

wait, where did the 10000 go?

Answer 11

\[(\log 39145)/(\log 1.0025) = 12t \]

Answer 12

You were close.
39,145(1+0.03/12)^12t = 10,000
try again with this!

Answer 13

So 39,145(1+0.03/12) is the base, and 10,000 is the argument?

Answer 14

\[39145\left( 1.0025 \right)^{12t}=10000\]
\[\frac{ 39145 }{ 10000 }=\left( 1.0025 \right)^{-12t}\]
\[\log 39,145-\log 10,000=-12t \log \left( 1.0025 \right)\]
\[t=-\frac{ \log 39,145-\log 10,000 }{12 \times \log 1.0025 }=?\]

Answer 15

Take log on both sides, like @surjithayer did.

Answer 16

Thanks guys!