How many years will it take an investment of 1000 to double if the interest rate is 6% compounded annually

Years = {log(total) -log(Principal)} ÷ log(1 + rate)

Years = log(2,000) - log(1,000) / log(1.06)

Years = 3.3010299957 - 3 / 0.025305865265

Years = .3010299957 / 0.025305865265

Years = 11.8956610473 or about 11.9 Years

Step-by-step explanation:

11.9 years

Step-by-step explanation:

The compound amount formula applicable here is A = P(1+r)^t.

Substituting the given data, $2000 = $1000(1.06)^t, or 2 = 1.06^t.

Taking the log of both sides, log 2 = t log 1.06.  Then t = 0.30103/0.02531, or t = 11.896.

The investment will doube in 11.9 years.

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step-by-step explanation:

du what else would it be

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y > 1/3x

y < = -x + 4

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