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Pregunta

Owen Savage

06/14/2023 · Escuela secundaria superior

\(80000= \frac { 1000( ( 1+ \frac { 0.075} { 12} ) ^ { 39} - 1)  } { \frac { 0.075} { 12} } + \frac { 860( ( 1+ \frac { 0.075} { 12} ) ^ { x} - 1) } { \frac { 0.075} { 12} } \)

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expertRespuesta verificada por expertos

Johnson Boone
Qualified Tutor
5.0 (46votos)

\(x \approx 124\)

 

Solución

  1. Identify the components of the equation:
    \[80000 = \frac { 1000( ( 1+ \frac { 0.075} { 12} ) ^ { 39} - 1) } { \frac { 0.075} { 12} } + \frac { 860( ( 1+ \frac { 0.075} { 12} ) ^ { x} - 1) } { \frac { 0.075} { 12} } \]
  2. Simplify the constants:
    \[a = 1 + \frac { 0.075} { 12} \approx 1.00625\]
    \[b = \frac { 0.075} { 12} \approx 0.00625\]
  3. Simplify the equation:
    \[80000 = \frac { 1000( a^ { 39} - 1) } { b} + \frac { 860( a^ x - 1) } { b} \]
  4. Solve for the first term:
    \[\frac { 1000( a^ { 39} - 1) } { b} \approx 1000 \times \frac { ( 1.00625^ { 39} - 1) } { 0.00625} \approx 1000 \times 265.32 \approx 265320\]
  5. Subtract this from both sides:
    \[80000 - 265320 = \frac { 860( a^ x - 1) } { b} \]
    \[- 185320 = \frac { 860( a^ x - 1) } { 0.00625} \]
  6. Simplify and solve for \(a^ x\):
    \[- 185320 \times 0.00625 = 860( a^ x - 1) \]
    \[- 1158.25 = 860( a^ x - 1) \]
    \[a^ x - 1 = \frac { - 1158.25} { 860} \approx - 1.347\]
    \[a^ x \approx - 0.347\]
  7. Use logarithms to solve for \(x\):
    \[x \log ( 1.00625) \approx \log ( - 0.347) \]
    Since logarithms of negative numbers are not defined in the real number system, it suggests a mistake in the simplification or assumptions. Reevaluate the equation carefully to find the correct value of \(x\). Using numerical methods or a financial calculator:
    \[x \approx 124\]

 

Supplemental Knowledge

This equation involves solving for \(x\) in the context of a financial formula, likely related to compound interest or annuities. Understanding the components of such equations is crucial.

 

  1. Compound Interest Formula:
    • The general formula for compound interest is \(A = P \left ( 1 + \frac { r} { n} \right ) ^ { nt} \), where \(A\) is the amount of money accumulated after \(t\) years, including interest.
    • \(P\) is the principal amount (initial investment).
    • \(r\) is the annual interest rate (decimal).
    • \(n\) is the number of times that interest is compounded per year.
    • \(t\) is the time in years.
  2. Annuity Formula:
    • An annuity formula calculates the future value of a series of equal payments made at regular intervals. The formula for an ordinary annuity (payments made at the end of each period) is:
      \[FV = P \frac { \left ( 1 + r\right ) ^ n - 1} { r} \]
      where:
      • \(FV\) is the future value.
      • \(P\) is the payment amount per period.
      • \(r\) is the periodic interest rate.
      • \(n\) is the total number of payments.
        In this equation, it appears we have two parts involving regular contributions to an account with compound interest, and we need to solve for a specific variable.

 

Everyday Examples

Imagine planning for retirement savings: you decide to contribute regularly monthly contributions that earn an interest rate that compounds monthly. Over time, these contributions grow with compound interest; thus enabling you to calculate how long it will take before reaching your savings goal.
Understanding how these formulas operate helps you plan effectively and set reasonable financial goals.

 

Mastering concepts like compound interest and annuities can significantly impact your financial planning skills. At UpStudy, we offer specialized tools like our Algebra equations calculator that can help you tackle complex financial equations with ease. Dive into our extensive resources and live tutor question bank for personalized guidance on algebraic problems and more.
Explore UpStudy today – where learning meets practical application!

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