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!
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Sincerely,
Piyush A.
