Question

How do you work out the rate, time and principal from the compound interest formula?


`A = P( {1 + (r)/(100) })^(n)`
​

Answer 1

Part 1)
`P=A/(1+(r)/(n))^(nt)` (see the explanation)

Part 2)
`r=n[(A)/(P)^(1/(nt))-1]` (see the explanation)

Part 3)
`t=log((A)/(P))/[(n)log(1+(r)/(n))]` (see the explanation)

Explanation:

we know that

The compound interest formula is equal to

`A=P(1+(r)/(n))^(nt)`

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part 1) Find the Principal P

The values of A,r,n and t are given

Isolate the variable P

`A=P(1+(r)/(n))^(nt)`

Divide both side by
`(1+(r)/(n))^(nt)`

`P=A/(1+(r)/(n))^(nt)`

Part 2) Find the rate r

The values of A,P,n and t are given

Isolate the variable r

`A=P(1+(r)/(n))^(nt)`

Divide both sides by P

`(A)/(P) =(1+(r)/(n))^(nt)`

Elevated both sides to 1/(nt)

`(A)/(P)^(1/(nt)) =(1+(r)/(n))`

subtract 1 both sides

`(A)/(P)^(1/(nt))-1 =(r)/(n)`

Multiply by n both sides

`r=n[(A)/(P)^(1/(nt))-1]`

Part 3) Find the time t

The values of A,P,r and n are given

Isolate the variable t

`A=P(1+(r)/(n))^(nt)`

Divide both sides by P

`(A)/(P) =(1+(r)/(n))^(nt)`

Apply log both sides

`log((A)/(P))=log(1+(r)/(n))^(nt)`

Apply property of exponents

`log((A)/(P))=(nt)log(1+(r)/(n))`

Divide both side by
`(n)log(1+(r)/(n))`

`t=log((A)/(P))/[(n)log(1+(r)/(n))]`