Question

How long does it take to double a $1,000 investment that pays 6.5% annual interest, compounded monthly?

Answer 1

Compound interest formula: A=P(1+interst)^n

P being the capital invested & n = the number of years, A the new capital

A= 1000(1+0.065)^n. They want to know n when the new capital is doubled,

2000=1000(1.065)^n & you will find n=11 years

Answer 2

It would take approximately 11 years to double a $1,000 investment that pays 6.5% annual interest rate and compounded on a monthly basis.

How to calculate compound interest?

Mathematically, compound interest is calculated by using this formula:

$A=P(1+\frac{r}{n} )^{nt}$

Where:

• A is the future value.

• P is the principal.

• R is the interest rate.

• T is the time measured in years.

• n is the number of times compounded.

Making t the subject of formula, we have:

$t = \frac{ln(A) - ln(P)}{n(ln(1 + \frac{r}{n} ))}$

Given the following data:

Principal = $1,000.

Future value = $2,000.

Interest rate = 6.5% = 0.065.

Number of times compounded = 12.

Substituting the given parameters into the formula, we have;

$t = \frac{ln(2000) - ln(1000)}{12(ln(1 + \frac{0.065}{12} ))} \\\\t = \frac{7.6 - 6.91}{12(ln(1.0054 ))} \\\\t=\frac{0.69}{12(0.00539} \\\\t=\frac{0.69}{0.06468}$

Time, t = 10.66 ≈ 11 years.

Read more on interest here: brainly.com/question/24341207