A sum of money at compound interest becomes twice the principal in 3 years. In how many years will it become 64 times its principal amount?

Given:

A sum of money becomes twice the principal in 3 years at compound interest.

Formula Used:

Compound Interest Formula: A = P (1 +\( \frac{r}{100}\))ⁿ

Where:

• A = Amount
• P = Principal
• r = Rate of interest per period
• n = Number of periods

Calculation:

Given, A = 2P and n = 3 years

⇒ 2P = P (1 + r/100)³

⇒ 2 = (1 + r/100)³

Taking cube root on both sides:

⇒ ∛(2) = 1 + r/100

⇒ 1.2599 = 1 + r/100

⇒ r/100 = 0.2599

⇒ r = 25.99%

Now, we need to find the time (n) for the amount to become 64 times the principal.

Given, A = 64P

⇒ 64P = P (1 + \(\frac{25.99}{100}\))ⁿ

⇒ 64 = (1 + 0.2599)ⁿ

⇒ 64 = 1.2599ⁿ

Taking logarithm on both sides:

⇒ log(64) = n log(1.2599)

⇒ log(2⁶) = n log(1.2599)

⇒ 6 log(2) = n log(1.2599)

⇒ n = \( \frac{6log(2)}{log(1.2599)}\)

Using logarithm values:

⇒ n = \( \frac{6 × 0.3010}{0.1004}\)

⇒ n = 18

The sum of money will become 64 times its principal amount in 18 years.