Question

I need help!

Answer 1

Answer:  23 years

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Work Shown:

P = 1000 is the amount deposited

We want this value to double to A = 2000 which is the amount in the account at time t (in years).

r = 0.03 represents the interest rate in decimal form.

The value of t is unknown but we can solve for it like so

$A = Pe^{rt}\\\\2000 = 1000e^{0.03t}\\\\2 = e^{0.03t}\\\\\ln(2) = \ln\left(e^{0.03t}\right)\\\\\ln(2) = 0.03t\ln\left(e\right)\\\\\ln(2) = 0.03t*1\\\\\ln(2) = 0.03t$

$0.03t = \ln(2)\\\\t = \frac{\ln(2)}{0.03}\\\\t \approx 23.1049060186649\\\\t \approx 23$

It will take about 23 years for the amount to double.

As a check,

$A = 1000e^{0.03t}\\\\A \approx 1000e^{0.03*23.104906}\\\\A \approx 1,999.9999988806\\\\A \approx 2000$

which helps show that after roughly 23 years, we'll have about 2000 dollars in the account.

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Side note: Use of the rule of 72 leads to 72/3 = 24. The '3' is from the 3% interest rate. So the rule of 72 says it will take about 24 years for the amount to double. This isn't too far off from the 23 answer we got.