Math / Calculus

QuestionYou deposit $\$ 3000$ in an account earning $7 \%$ interest compounded continuously. The amount of money in the account after $t$ years is given by $A(t)=3000 e^{0.07 t}$ .
How much will you have in the account in 5 years? \\qquad$ Round your answer to 2 decimal places.
How long will it be until you have $\$ 17600$ in the account? $\square$ years. Round your answer to 2 decimal places.
How long does it take for the money in the account to double? $\square$ years. Round your answer to 2 decimal places.

Studdy Solution

STEP 1

1. The initial deposit is $3000.
2. The interest rate is $7\%$ compounded continuously.
3. The formula for continuous compounding is $A(t) = P e^{rt}$, where $P$ is the principal, $r$ is the rate, and $t$ is time in years.

STEP 2

1. Calculate the amount in the account after 5 years.
2. Determine the time required for the account balance to reach $17600.
3. Calculate the time required for the initial deposit to double.

STEP 3

Substitute $t = 5$ into the formula:
$A(5) = 3000 e^{0.07 \times 5}$

STEP 4

Calculate the exponent:
$A(5) = 3000 e^{0.35}$

STEP 5

Calculate $e^{0.35}$ using a calculator:
$e^{0.35} \approx 1.4190675$

STEP 6

Calculate $A(5)$ :
$A(5) = 3000 \times 1.4190675 \approx 4257.20$
So, the amount in the account after 5 years is approximately $\$4257.20$ .

STEP 7

Set the equation for $A(t) = 17600$ :
$17600 = 3000 e^{0.07 t}$

STEP 8

Solve for $e^{0.07 t}$ :
$\frac{17600}{3000} = e^{0.07 t}$ $\frac{17600}{3000} \approx 5.8667$

STEP 9

Take the natural logarithm of both sides:
$\ln(5.8667) = 0.07 t$

STEP 10

Calculate $\ln(5.8667)$ using a calculator:
$\ln(5.8667) \approx 1.768$

STEP 11

Solve for $t$ :
$t = \frac{1.768}{0.07} \approx 25.26$
So, it will take approximately $25.26$ years to have $\$17600$ in the account.

STEP 12

Set the equation for doubling the initial deposit:
$6000 = 3000 e^{0.07 t}$

STEP 13

Solve for $e^{0.07 t}$ :
$\frac{6000}{3000} = e^{0.07 t}$ $2 = e^{0.07 t}$

STEP 14

Take the natural logarithm of both sides:
$\ln(2) = 0.07 t$

STEP 15

Calculate $\ln(2)$ using a calculator:
$\ln(2) \approx 0.693$

STEP 16

Solve for $t$ :
$t = \frac{0.693}{0.07} \approx 9.90$
So, it will take approximately $9.90$ years for the money in the account to double.
The amount in the account after 5 years is approximately $\$4257.20$ .
The time until the account balance reaches $\$17600$ is approximately $25.26$ years.
The time it takes for the money to double is approximately $9.90$ years.

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Studdy Solution

STEP 1

1. The initial deposit is $3000.
2. The interest rate is \(7\%\) compounded continuously.
3. The formula for continuous compounding is \(A(t) = P e^{rt}\), where \(P\) is the principal, \(r\) is the rate, and \(t\) is time in years.

STEP 2

1. Calculate the amount in the account after 5 years.
2. Determine the time required for the account balance to reach $17600.
3. Calculate the time required for the initial deposit to double.

STEP 3

Substitute $t = 5$ into the formula:
$A(5) = 3000 e^{0.07 \times 5}$

STEP 4

Calculate the exponent:
$A(5) = 3000 e^{0.35}$

STEP 5

Calculate $e^{0.35}$ using a calculator:
$e^{0.35} \approx 1.4190675$

STEP 6

Calculate $A(5)$ :
$A(5) = 3000 \times 1.4190675 \approx 4257.20$
So, the amount in the account after 5 years is approximately $\$4257.20$ .

STEP 7

Set the equation for $A(t) = 17600$ :
$17600 = 3000 e^{0.07 t}$

STEP 8

Solve for $e^{0.07 t}$ :
$\frac{17600}{3000} = e^{0.07 t}$ $\frac{17600}{3000} \approx 5.8667$

STEP 9

Take the natural logarithm of both sides:
$\ln(5.8667) = 0.07 t$

STEP 10

Calculate $\ln(5.8667)$ using a calculator:
$\ln(5.8667) \approx 1.768$

STEP 11

Solve for $t$ :
$t = \frac{1.768}{0.07} \approx 25.26$
So, it will take approximately $25.26$ years to have $\$17600$ in the account.

STEP 12

Set the equation for doubling the initial deposit:
$6000 = 3000 e^{0.07 t}$

STEP 13

Solve for $e^{0.07 t}$ :
$\frac{6000}{3000} = e^{0.07 t}$ $2 = e^{0.07 t}$

STEP 14

Take the natural logarithm of both sides:
$\ln(2) = 0.07 t$

STEP 15

Calculate $\ln(2)$ using a calculator:
$\ln(2) \approx 0.693$

STEP 16

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Studdy Solution

STEP 1

1. The initial deposit is $3000.2. The interest rate is \(7\%\) compounded continuously.3. The formula for continuous compounding is \(A(t) = P e^{rt}\), where \(P\) is the principal, \(r\) is the rate, and \(t\) is time in years.

STEP 2

1. Calculate the amount in the account after 5 years.2. Determine the time required for the account balance to reach $17600.3. Calculate the time required for the initial deposit to double.

STEP 3

Substitute t=5t = 5t=5 into the formula:A(5)=3000e0.07×5 A(5) = 3000 e^{0.07 \times 5} A(5)=3000e0.07×5

STEP 4

Calculate the exponent:A(5)=3000e0.35 A(5) = 3000 e^{0.35} A(5)=3000e0.35

STEP 5

Calculate e0.35e^{0.35}e0.35 using a calculator:e0.35≈1.4190675 e^{0.35} \approx 1.4190675 e0.35≈1.4190675

STEP 6

Calculate A(5)A(5)A(5):A(5)=3000×1.4190675≈4257.20 A(5) = 3000 \times 1.4190675 \approx 4257.20 A(5)=3000×1.4190675≈4257.20So, the amount in the account after 5 years is approximately $4257.20\$4257.20$4257.20.

STEP 7

Set the equation for A(t)=17600A(t) = 17600A(t)=17600:17600=3000e0.07t 17600 = 3000 e^{0.07 t} 17600=3000e0.07t

STEP 8

Solve for e0.07te^{0.07 t}e0.07t:176003000=e0.07t \frac{17600}{3000} = e^{0.07 t} 300017600​=e0.07t 176003000≈5.8667 \frac{17600}{3000} \approx 5.8667 300017600​≈5.8667

STEP 9

Take the natural logarithm of both sides:ln⁡(5.8667)=0.07t \ln(5.8667) = 0.07 t ln(5.8667)=0.07t

STEP 10

Calculate ln⁡(5.8667)\ln(5.8667)ln(5.8667) using a calculator:ln⁡(5.8667)≈1.768 \ln(5.8667) \approx 1.768 ln(5.8667)≈1.768

STEP 11

Solve for ttt:t=1.7680.07≈25.26 t = \frac{1.768}{0.07} \approx 25.26 t=0.071.768​≈25.26So, it will take approximately 25.2625.2625.26 years to have $17600\$17600$17600 in the account.

STEP 12

Set the equation for doubling the initial deposit:6000=3000e0.07t 6000 = 3000 e^{0.07 t} 6000=3000e0.07t

STEP 13

Solve for e0.07te^{0.07 t}e0.07t:60003000=e0.07t \frac{6000}{3000} = e^{0.07 t} 30006000​=e0.07t 2=e0.07t 2 = e^{0.07 t} 2=e0.07t

STEP 14

Take the natural logarithm of both sides:ln⁡(2)=0.07t \ln(2) = 0.07 t ln(2)=0.07t

STEP 15

Calculate ln⁡(2)\ln(2)ln(2) using a calculator:ln⁡(2)≈0.693 \ln(2) \approx 0.693 ln(2)≈0.693

STEP 16

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1. The initial deposit is $3000.
2. The interest rate is \(7\%\) compounded continuously.
3. The formula for continuous compounding is \(A(t) = P e^{rt}\), where \(P\) is the principal, \(r\) is the rate, and \(t\) is time in years.

1. Calculate the amount in the account after 5 years.
2. Determine the time required for the account balance to reach $17600.
3. Calculate the time required for the initial deposit to double.

Substitute $t = 5$ into the formula:
$A(5) = 3000 e^{0.07 \times 5}$

Calculate the exponent:
$A(5) = 3000 e^{0.35}$

Calculate $e^{0.35}$ using a calculator:
$e^{0.35} \approx 1.4190675$

Calculate $A(5)$ :
$A(5) = 3000 \times 1.4190675 \approx 4257.20$
So, the amount in the account after 5 years is approximately $\$4257.20$ .

Set the equation for $A(t) = 17600$ :
$17600 = 3000 e^{0.07 t}$

Solve for $e^{0.07 t}$ :
$\frac{17600}{3000} = e^{0.07 t}$ $\frac{17600}{3000} \approx 5.8667$

Take the natural logarithm of both sides:
$\ln(5.8667) = 0.07 t$

Calculate $\ln(5.8667)$ using a calculator:
$\ln(5.8667) \approx 1.768$

Solve for $t$ :
$t = \frac{1.768}{0.07} \approx 25.26$
So, it will take approximately $25.26$ years to have $\$17600$ in the account.

Set the equation for doubling the initial deposit:
$6000 = 3000 e^{0.07 t}$

Solve for $e^{0.07 t}$ :
$\frac{6000}{3000} = e^{0.07 t}$ $2 = e^{0.07 t}$

Take the natural logarithm of both sides:
$\ln(2) = 0.07 t$

Calculate $\ln(2)$ using a calculator:
$\ln(2) \approx 0.693$