Question

chelsea put $7500 into an account paying 5% compounded continuously. she now has $10,643.01. how long has the money been in the account?

Answer 1

Answer: 7.17358353 or just 7.17

Step-by-step explanation:

Firstly to solve this problem we must set up an equation which should look similar to this:

P(t) = 7500(1.05)^t

The percentage value would be 1.05 since it is increasing, and they also give us the total as well, what we just need now is time.

10,643.01 = 7500(1.05)^t

We must get rid of the 7500 so we divide both sides by 7500:

1.419068 = (1.05)^t

We now must use log to isolate t on one side of the equation:

log₁.₀₅(1.419068) = t

Unless you have a super calculator, you can plug this in the calculator and get the answer but we can just use the natural log (ln) to make this easier.

ln(1.419068) / ln(1.05) = t

t = 7.17358353

Generally we round up to the hundredths place for problems asking for time so a better estimated answer would just be 7.17

Answer 2

To find out how long the money has been in the account, we can use the formula for continuously compounded interest. Substituting the given values into the formula and solving for t, we find that the money has been in the account for approximately 4.5 years.

Step-by-step explanation:

To find out how long the money has been in the account, we can use the formula for continuously compounded interest:

A = P*e^(rt)

Where:

• A = Final amount of money in the account ($10,643.01)
• P = Initial amount of money in the account ($7500)
• r = Interest rate (5% or 0.05)
• t = Time in years

Substituting the given values into the formula, we have:

10,643.01 = 7500*e^(0.05t)

To solve for t, we need to isolate it on one side of the equation. Divide both sides of the equation by 7500:

e^(0.05t) = 10,643.01 / 7500

Next, take the natural logarithm of both sides to eliminate the exponential:

0.05t = ln(10,643.01 / 7500)

Finally, divide both sides of the equation by 0.05 to solve for t:

t = ln(10,643.01 / 7500) / 0.05

Using a calculator, we can find the value of t to be approximately 4.5 years.