Answer:
10.058 years for $300 to grow to $9600 at the same interest rate.Explanation:
To solve this problem, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A is the final amount
P is the principal amount (initial amount)
r is the annual interest rate (as a decimal)
n is the number of times the interest is compounded per year
t is the number of years
For the first scenario:
P = $5000
A = 2P = $10000
t = 6 years
We need to find the interest rate, r, that will make the amount double in 6 years. Let's substitute the values into the formula and solve for r:
$10000 = $5000(1 + r/1)^(1*6)
2 = (1 + r)^6
Taking the sixth root of both sides:
1 + r = √2
r = √2 - 1
Now we can use this interest rate to solve for the time it will take for $300 to grow to $9600.
P = $300
A = $9600
r = √2 - 1
Let's substitute the values into the formula and solve for t:
$9600 = $300(1 + (√2 - 1)/1)^(1*t)
32 = (√2)^(t)
Taking the logarithm of both sides (with base √2):
log(32) = log(√2)^t
log(32) = t * log(√2)
Using a calculator to evaluate the logarithms:
t ≈ log(32) / log(√2)
t ≈ 10.058
Therefore, it will take approximately 10.058 years for $300 to grow to $9600 at the same interest rate.