Compound Interest and Exponential Functions

Question

How much money should you put in a savings account now that earns 5% a year compounded daily if you want to have $32,000 in 18 years? 

 

Would my formula be right going off the model A=P(1+r/n)^nt where

 

A=amount of time t

P=principal

r=interest rate

t=time in years

n=number of times compounded in a year.

 

32,000=P(1+(.05/365))^6570

 

...and then how do you find the answer? Please include all steps and why you did it so that I may understand the whole process. Thank you!

 

Megan

Answer 1

Hi, Megan,

 

Your equation looks perfect to me - very nice job!

 

The only "technically" that I can think of is that every 4 years there will be a leap year, in which there will be 366 days in a year - but I think that is a minor point, and if you specifically state that you are assuming each year has 365 days, this is certainly good enough.

 

Now, you just need to solve for P.

 

    32000 = P (1 + .05/365)⁶⁵⁷⁰

 

P is being multiplied by the exponential term, so in order to "undo" that multiplication, we will divide both sides of the equation by the exponential term. This will cancel it on the right hand side, leaving only P:

 

     32000                        = P

       (1 + .05/365)⁶⁵⁷⁰

 

     P = 13011.031

 

So you would need to have $13011.04 in order to have at least $32000 after 18 (non-leap) years. I hope this helps!

Answer 2

You have it set up perfectly!

 

32000=P(1+(.05/365))^6570

 

Now just solve for P:

 

P = 32000 / ( (1+(.05/365))^6570 )

 

Now you need to use a calculator. Every calculator may be different, so you need your own and know how to use it.

 

[button] will be my notation for a key on my calculator; so here's how I would do it:

 

0.05 [/] 365 [+] 1 [=] [y^x] 6570 [=] [1/x] [*] 32000 [=]

and the calculator shows: 13011.031063608

 

Which you'd probably round to the nearest cent: $13,011.03.

 

Answer 3

 A = P ( 1 + r/n) ^ n

 

 32000 = P ( 1 +0.05/365) ^( 365* 18)=

 

  P = 32000          ___               

         ( 1+ 0.05/365) ^ (365*18)

 

 

    log P = log ( 32000) - 365 *18 log ( 1 + 0.05/365)

 

  This is done by use of Log, directly is done by others.

  log P = 4.114

 

 

   P = 10 ^4.114 = 13001.7

 

      Testing the

 

        answer

 

      ( 1 + 0.05/ 365) ^( 18 * 365) = 2.46 / number of times principal grows in 18 years, under the

                                                     condition given

 

      32000 / 2.46  = 13008.13 / approximately.

Answer 4

0.05/365=0.0001369863

$32,000=P(1.0001369863)^6570

1.0001369863^6570=2.45945148817

32,000=P(2.45945)

32,000/2.45945=P

$13,011.03=P

There are websites that can do calculations with numbers having many digits and numbers involving large exponents.