Question

How long does it take for an investment to double in value if it is invested at 15?% compounded monthly??Compounded? continuously?

Answer 1

We know that the formula for componuding = A (1+i)ⁿ, let the amount invested be $ 1, it will become $2 in "n" number of months. i = monthly inerest rate = 15% / 12 = 1.25% or 0.0125,

and the formula will be $2 = $1 X (1.00125ⁿ), applying logarithm in both sides then we will get

log 2 = n log 1.0125.

The value of log 2 is 0.301029995663981 and value of log 1.0125 = 0.00539503188670614,

So 0.301029995663981 = 0.00539503188670614n,

n = 0.301029995663981 / 0.00539503188670614,

n = 55.7976304840286,

So 55.7976304840286 months will take to get $1 be $2, means if $ 1 invested for 55.7976304840286 months, we will get $2.

means 4.64980254033572 years or 4 years 7 months and 23.9289145208592 days.

If continuously Compounded,

Formula for continuously Compounding is e^rtX Amount invested, if it becomes 2,

the equation to solve this will be 2 = e^rt, where r is the interest rate = 15% or 0.15, t = time period.

approximate value of 'e' = 2.71828182845904.

Apply logarithm in both side of Equation then

log (2) = log (e^0.15t),

The value of log (2) is 0.301029995663981,

log (e^0.15t) = 0.15t log e,

log e = 0.434294482, then applying value in equation, we will get

0.301029995663981 = 0.15t (0.434294482),

then 0.15t = 0.301029995663981 / 0.434294482,

0.15t = 0.693147180405532,

t = 0.693147180405532 / 0.15,

t = 4.62098120270355,

that is 4 years, 7 months and days.

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