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Carlene is saving her money to buy a [tex]$\$[/tex]500[tex]$ desk. She deposits $[/tex]\[tex]$400$[/tex] into an account with an annual interest rate of [tex]$6\%$[/tex] compounded continuously. The equation [tex]$400 e^{0.06 t} = 500$[/tex] represents the situation, where [tex]$t$[/tex] is the number of years the money needs to remain in the account.

About how long must Carlene wait to have enough money to buy the desk? Use a calculator and round your answer to the nearest whole number.

A. 4 years
B. 7 years
C. 9 years
D. 20 years


Sagot :

To solve the problem of how long Carlene must wait to have enough money to buy the desk, we start with the given equation that represents the situation:

[tex]\[ 400 e^{0.06 t} = 500 \][/tex]

Here, [tex]\( 400 \)[/tex] is the initial deposit, [tex]\( 0.06 \)[/tex] is the annual interest rate, [tex]\( t \)[/tex] is the number of years, and [tex]\( 500 \)[/tex] is the desired amount Carlene wants to have.

We'll solve this equation step-by-step for [tex]\( t \)[/tex].

1. Isolate the exponential term:

Divide both sides of the equation by [tex]\( 400 \)[/tex]:

[tex]\[ e^{0.06 t} = \frac{500}{400} \][/tex]

[tex]\[ e^{0.06 t} = 1.25 \][/tex]

2. Apply the natural logarithm ([tex]\( \ln \)[/tex]) to both sides to solve for [tex]\( t \)[/tex]:

[tex]\[ \ln(e^{0.06 t}) = \ln(1.25) \][/tex]

3. Simplify the left side using the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex]:

[tex]\[ 0.06 t = \ln(1.25) \][/tex]

4. Solve for [tex]\( t \)[/tex]:

[tex]\[ t = \frac{\ln(1.25)}{0.06} \][/tex]

Using a calculator to evaluate [tex]\( \ln(1.25) \)[/tex]:

[tex]\[ \ln(1.25) \approx 0.223143551 \][/tex]

Now, substitute this value back into the equation:

[tex]\[ t = \frac{0.223143551}{0.06} \][/tex]

[tex]\[ t \approx 3.719059188 \][/tex]

5. Round to the nearest whole number:

[tex]\[ t \approx 4 \][/tex]

Therefore, Carlene must wait approximately 4 years to have enough money to buy the desk.

The correct answer is 4 years.