$$ \$ 600 $$ were deposited into an account wit a $$ 5 \% $$ interest rate, compounded continuously. How many years was it in the bank if the current amount is $$ \$ 6000 $$ ? $$ t=[?] \text { years } $$ Round to the nearest year.

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To solve this problem, we can use the formula for continuous compounding of interest:

$$ A = P e^{rt} $$

where:
- $$ A $$ is the current amount in the account,
- $$ P $$ is the initial deposit,
- $$ r $$ is the interest rate,
- $$ t $$ is the time in years.

Given:
- Initial deposit $$ P = \$600 $$,
- Interest rate $$ r = 5\% = 0.05 $$,
- Current amount $$ A = \$6000 $$.

We need to find the time $$ t $$ in years.

Substitute the given values into the formula:

$$ 6000 = 600 e^{0.05t} $$

Now, we can solve for $$ t $$ using the formula:

$$ t = \frac{\ln\left(\frac{6000}{600}\right)}{0.05} $$

Let's calculate the value of $$ t $$ using the formula.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\ln{\left(\frac{6000}{600}\right)}}{0.05}$$
- step1: Divide the terms:
  $$\frac{\ln{\left(10\right)}}{0.05}$$
- step2: Rewrite the expression:
  $$\frac{\ln{\left(10\right)}}{\frac{1}{20}}$$
- step3: Multiply by the reciprocal:
  $$\ln{\left(10\right)}\times 20$$
- step4: Reorder the terms:
  $$20\ln{\left(10\right)}$$
The time $$ t $$ in years is approximately 46.05 years.

Therefore, the account was in the bank for approximately 46 years.
The account was in the bank for approximately 46 years.

were deposited into an account wit
a interest rate, compounded
continuously. How many years was it in
the bank if the current amount is ?

Round to the nearest year.

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were deposited into an account wit a interest rate, compounded continuously. How many years was it in the bank if the current amount is ? Round to the nearest year.