If you deposited $1,000 in an account paying 6% interest compounded annually, how long would it take to double? Use formula A=P(1+r/n)^nt
+23254 Using that formula: A = 2000 P = 1000 r = 0.06 n= 1 Solve for t.
2000 = 1000(1 + 0.06/1)^(1·t)
2000 = 1000(1.06)^t
Divide by 1000:
2 = (1.06)^t
If the variable is an exponent, use logs:
log(2) = log( 1.06^t )
An exponent in a log comes out as a multiplier:
log(2) = t·log(1.06
Divide both sides by log(1.06):
log(2) / log(1.06) = t
t = 11.9 years (approximately)
+23254 Using that formula: A = 2000 P = 1000 r = 0.06 n= 1 Solve for t.
2000 = 1000(1 + 0.06/1)^(1·t)
2000 = 1000(1.06)^t
Divide by 1000:
2 = (1.06)^t
If the variable is an exponent, use logs:
log(2) = log( 1.06^t )
An exponent in a log comes out as a multiplier:
log(2) = t·log(1.06
Divide both sides by log(1.06):
log(2) / log(1.06) = t
t = 11.9 years (approximately)
