Pregunta
upstudy study bank question image url

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

Ask by Welch Norton. in the United States
Jan 21,2025

Solución de inteligencia artificial de Upstudy

Respuesta verificada por el tutor

Responder

It took approximately 46 years for the $600 to grow to $6000 with continuous compounding at a 5% interest rate.

Solución

To solve this problem, we can use the formula for continuous compounding of interest:
where:
  • is the current amount in the account,
  • is the initial deposit,
  • is the interest rate,
  • is the time in years.
Given:
  • Initial deposit ,
  • Interest rate ,
  • Current amount .
We need to find the time in years.
Substitute the given values into the formula:
Now, we can solve for using the formula for continuous compounding of interest.
Solve the equation by following steps:
  • step0: Solve for :
  • step1: Convert the expressions:
  • step2: Swap the sides:
  • step3: Divide both sides:
  • step4: Divide the numbers:
  • step5: Take the logarithm of both sides:
  • step6: Evaluate the logarithm:
  • step7: Multiply by the reciprocal:
  • step8: Multiply:

    The solution to the equation is .
Now, we can calculate the value of by substituting the value of into the equation.
Rounding to the nearest year, the time is approximately 46 years.

Respondido por UpStudy AI y revisado por un tutor profesional

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Extra Insights

To solve this, we can use the formula for continuous compounding:
Where:
  • is the amount of money accumulated after time (the final amount),
  • is the principal amount (the initial deposit),
  • is the annual interest rate (as a decimal),
  • is the time in years,
  • is the base of the natural logarithm (approximately equal to 2.71828).
Given:
Now plug in the values:
First, divide both sides by 600:
Next, take the natural logarithm of both sides:
Now, solve for :
Calculate (approximately 2.3026):
Rounding to the nearest year, the time is approximately:

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