Each period's interest joins the principal, so the next period earns on more.
Find the amount
₱20,000 is invested at 5% compounded annually for 3 years. What is it worth?
A = 20,000(1.05)³ = 20,000 × 1.157625 = ₱23,152.50. A is the simple-interest answer, ₱1,000 a year for three years — and the ₱152.50 gap between them is precisely the interest that the interest earned. C is one year only.
The second year earned ₱1,236, not ₱1,200. That extra ₱36 is 3% of the first year's interest — interest on interest, which is the whole idea.
A = P × (1 + r)ⁿ
the whole formula
1 + r keeps the balance and adds the growth
5% gives 1.05
n counts the periods
3 years, compounded yearly
I = A − P
interest is what is left over
Multiplying by 0.05 each year would leave 5% of the money, not add 5% to it. The 1 in (1 + r) is the part you already had, and it has to be there.
Over one period the two are identical. The curve only pulls away once there is interest for the interest to act on, which is why short loans often quote simple interest and long investments never do.
A = P × (1 + r)ⁿ
A is known, P is not
60000 = P(1.05)⁴
substitute
P = 600001.21550625
divide, do not subtract
P ≈ ₱49,362.15
the present value
This is what "the time value of money" means in practice: ₱49,362 today and ₱60,000 in four years are the same thing at 5%.
When n is the unknown, it is the exponent itself, so we divide by the principal and take logs of both sides.
How long does ₱8,000 take to grow to ₱12,000 at 5% compounded annually? (log 1.5 ≈ 0.1761, log 1.05 ≈ 0.0212)
12000 = 8000(1.05)ⁿ
substitute
1.5 = (1.05)ⁿ
divide by the principal
log 1.5 = n · log 1.05
the power law brings n down
n = 0.1761 ÷ 0.0212 ≈ 8.3
about 8.3 years
Interest is added only once a year, so the balance first reaches ₱12,000 at the end of year 9: 8,000(1.05)⁸ ≈ ₱11,819.64, still short, and 8,000(1.05)⁹ ≈ ₱12,410.63.
Three years, so a cube root: ₱8,000 grows to ₱10,648 in 3 years.
106488000 = 1.331 = (1 + r)³
divide by the principal
1 + r = 1.1
cube root: 1.1 × 1.1 × 1.1 = 1.331
r = 10%
subtract the 1
The root matches the number of periods: square root for 2, cube root for 3.
A value that shrinks: a ₱80,000 motorbike loses 10% a year, compounded.
base 1 − 0.10 = 0.9
keeps 90% each year
A = 80000(0.9)³ = 80000 × 0.729
three years
A = ₱58,320
not ₱56,000
Taking 30% off once gives ₱56,000. Each year's 10% comes off a smaller value, so the loss is less.
A fractional rate: ₱16,000 at 2.5% compounded annually for 2 years
base 1.025
2.5% is 0.025
(1.025)² = 1.050625
square the base
A = 16000 × 1.050625 = ₱16,810
interest ₱810
Keep every decimal of the base until the last line; rounding 1.025 to 1.03 would give ₱16,974.40.
28090 = 25000(1 + r)²
r is the unknown
1.1236 = (1 + r)²
divide by the principal
1.06 = 1 + r
square root of both sides
r = 0.06, so 6%
subtract the 1 back off
The unknown decides the tool. An unknown rate sits inside a power, so take a root. An unknown time is the exponent itself, so take logarithms. Reaching for the wrong one is the usual reason these problems stall.
Periodical exams ask for the amount or the interest after a few years, or the principal needed for a target; entrance tests like the unknown rate (take a root) and the comparison with simple interest. Write A = P × (1 + r)ⁿ, decide whether the unknown is A, P, r or n, and undo accordingly. For a 10-second check, the compound amount must be a little MORE than the simple-interest amount for the same P, r and time.
Find the interest earned
₱30,000 at 6% compounded annually for 2 years. How much INTEREST is earned?
A = 30,000(1.06)² = 30,000 × 1.1236 = ₱33,708, so the interest is 33,708 − 30,000 = ₱3,708. A is the simple-interest figure, and the ₱108 difference is the second year's interest on the first year's ₱1,800. D is the amount, not the interest.
Hint: work out the amount first, then subtract the principal. The formula gives A; the question wants I.
