How long would it take money deposited in an account earning 2.8% per year, compounded quarterly, to double?
+23254 The compound interest formula is: A = P(1 + r/n)^(n·t)
where A = final amount P = amount invested r = rate (as a decimal)
n = number of times compounded per year t = number of years
You can choose any number that you want for P; I will choose 1. (This procedure works for any number that you choose for P.) The value of A is twice that of P, so A = 2 r = .028 n = 4
Substituting: 2 = 1(1 + .028/4)^(4t) ---> 2 = (1.007)^(4t)
Since your unknown is in an exponent, use logs: log(2) = log[(1.007)^4t]
An exponent in a log comes out as a multiplier: log(2) = 4t·log(1.007)
log(2) / log(1.007) = 4t
t = [ log(2) / log(1.007) ] ÷ 4
t ≈ 24.8 years
+23254 The compound interest formula is: A = P(1 + r/n)^(n·t)
where A = final amount P = amount invested r = rate (as a decimal)
n = number of times compounded per year t = number of years
You can choose any number that you want for P; I will choose 1. (This procedure works for any number that you choose for P.) The value of A is twice that of P, so A = 2 r = .028 n = 4
Substituting: 2 = 1(1 + .028/4)^(4t) ---> 2 = (1.007)^(4t)
Since your unknown is in an exponent, use logs: log(2) = log[(1.007)^4t]
An exponent in a log comes out as a multiplier: log(2) = 4t·log(1.007)
log(2) / log(1.007) = 4t
t = [ log(2) / log(1.007) ] ÷ 4
t ≈ 24.8 years
