Solving for n: 5000=200(1-(1.01)^-n/.01

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In summary, the equation "5000=200(1-(1.01)^-n/.01" is an exponential decay function used to calculate the present value of a future sum of money with a discount rate of 1%. To solve for n, algebraic manipulation and logarithms can be used. The number 1.01 represents the growth rate or interest rate, and this equation can be applied to various real-life scenarios, such as finance, economics, and financial planning.
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Homework Statement


5000=200(1-(1.01)to the -n power, divided by .01


Homework Equations





The Attempt at a Solution

 
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Take logs of both sides.
 
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FAQ: Solving for n: 5000=200(1-(1.01)^-n/.01

What does the equation "5000=200(1-(1.01)^-n/.01" mean?

This equation is known as an exponential decay function, where n represents the number of time periods. It is used to calculate the present value of a future sum of money, with a discount rate of 1%. The goal is to solve for n, which represents the number of time periods required for the present value to equal 5000.

How do I solve for n in this equation?

To solve for n, you can use algebraic manipulation and logarithms. Start by dividing both sides of the equation by 200, then take the natural logarithm of both sides. Using the power rule, you can then isolate n and solve for it.

What is the significance of the number 1.01 in this equation?

The number 1.01 represents the growth rate or interest rate. Since the equation has a discount rate of 1%, the growth rate will be 1% less, which is equivalent to 0.01. This value is used to calculate the decrease in value over each time period.

Can this equation be used for any type of problem?

Yes, this equation is a general form of the present value formula and can be used for any problem that involves exponential decay with a discount rate of 1%. It is commonly used in finance and economics to evaluate investments and loans.

What are some real-life applications of this equation?

This equation can be applied to various real-life scenarios, such as calculating the present value of a loan with a 1% interest rate, determining the depreciation of an asset with a 1% decrease per year, or evaluating the growth of a population with a 1% decrease in birth rate. It can also be used in financial planning to determine the amount of money needed to save in order to reach a future goal, taking into account inflation and interest rates.

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