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A person invested $7,400 in an account growing at a rate allowing the money todouble every 6 years. How much money would be in the account after 4 years, to thenearest dollar?

User Stargazer
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1 Answer

16 votes
16 votes

The rate at which the money is growing is exponential. The formula for determining such exponential growth is expressed as

A(t) = Pe^rt

Where

A(t) represents the amount after t years

P represents the initial amount

t represents the number of years

r represents the growth rate

Given that the amount doubles after 6 years, it means that

when t = 6, A(t) = 7400 * 2 = 14800

Recall, P = 7400

Thus, the expression becomes

14800 = 7400e^6t

14800/7400 = e^6t

2 = e^6t

Taking natural logarithm of both sides of the equation, it becomes

ln 2 = ln e^6t = 6t lne

Recall, ln e = 1

Thus, we have

ln 2 = 6t

t = ln2/6 = 0.1155

The equation would be

A(t) = 7400e^0.1155t

When t = 4, it becomes

A(t) = 7400e^0.1155*4

A(t) = 11745.62

Rounding to the nearest dollar, the amount in the account after 4 years is

$11746

User Doug Breaux
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