Vincelauret wrote:An investment of $1,500 was made in a certain bank account and it earned interest that was compounded annually; the annual interest rate was fixed for the entire duration of the investment. If after 12 years the $1,500 increased to $24,000 by earning interest, in how many years after the initial investment was made would the $1,500 have increased to $96,000 by earning interest at the same rate?
A. 15
B. 18
C. 20
D. 21
E. It cannot be determined from the information given

I found the answer by testing the different options, but does it exist an algebric formula?

We could use the formula for EXPONENTIAL GROWTH:

Final amount = (original amount) * (multiplier)^(number of changes).

After 12 years the $1,500 increased to $24,000.
In this case:
Final amount = 24000
Original amount = 1500
Multiplier = m
Number of changes = 12.

Plugging these values in the formula, we get:
24000 = 1500 * m¹²
16 = m¹²
(2�)^(1/12) = m
2^(1/3) = m.

How many years after the initial investment was made would the $1,500 have increased to $96,000?
In this case:
Final amount = 96000
Original amount = 1500
Multiplier = 2^(1/3)
Number of changes = x.

Plugging these values in the formula, we get:
96000 = 1500 * (2^1/3)^x
64 = 2^(x/3)
2� = 2^(x/3).
Since the bases are equal, the exponents must also be equal.
Thus:
x/3 = 6
x = 18.

The correct answer is B.

Vincelauret wrote:An investment of $1,500 was made in a certain bank account and it earned interest that was compounded annually; the annual interest rate was fixed for the entire duration of the investment. If after 12 years the $1,500 increased to $24,000 by earning interest, in how many years after the initial investment was made would the $1,500 have increased to $96,000 by earning interest at the same rate?
A. 15
B. 18
C. 20
D. 21
E. It cannot be determined from the information given

Slight variation of Brent's approach:

After 12 years the $1,500 increased to $24,000.
24000/1500 = 16.
Since 16 = 2�, the implication is that 12 YEARS are required for the investment to DOUBLE 4 TIMES.

How many years after the initial investment was made would the $1,500 have increased to $96,000?
96000/1500 = 64.
Since 64 = 2�, we must determine HOW MANY YEARS are required for the investment to DOUBLE 6 TIMES.

Thus, we could set up the following proportion:
12/4 = x/6
3 = x/6
x = 18.

The correct answer is B.

After painfully trying to solve this equation, no one had a good solution out there using the formula!

I finally figured it out using a structured easy to remember method. I have a more structured way to solve this problem using exponents and simultaneous equations.

Here it is.

Compound interest formula = P(1+r/n)^nt

P = initial amount
r = interest rate
n = number of times the interest is applied in a year
t = number of periods in question

Here's what know:
1. P = 1,500 since that's what's originally invested
2. The first relationship tells us that the resulting investment yields 24,000, so the first relationship will use the equation with 24,000.
3. The second relationship tells us that the resulting investment yields 96,000, so the second relationship will use the equation with 96,000.
n = 1 since the story tells us that the interest was compounded "annual" (instead of bi-annually, etc...)

(1) 24,000 = 1,500*(1+r/1)^(1*12)
(2) 96,000 = 1500*(1+r/1)^(1*t)

Let's solve for t.

Steps:
1. Simplify (1) and (2):
(1) 16 = (1+r)^12
(2) 64 = (1+r)^t

2. Let's put (1+r) to one side in (1), so: 12th_root(16) = 1+r, which is also the same as 1+r = 16^(1/12)
3. Now, let's plug in 1+r into (2) and solve for t.
When you plug 1+r into (2), you will realize that exponent (1/12) will multiply into t to give you t/12
64 = 16*[(1/12)*t] = 16^(t/12)
4. Now, remember that to solve for t, the bases must be the same so you can drop the bases completely like this:

64 = 4^3
16 = 4^2
Therefore:
4^3 = 4^[2*(t/12)] = 4^(2t/12) = 4^(t/6)
4^3 = 4^(t/6)

5. Finally, since the bases are 4s, drop the 4s and solve the equation:
3 = t/6
t = 18.

Answer is 18!