Compounding depreciation

Question

If something is worth 2,702 units on Day 1 and 2,316 units on Day X and is decreasing at a 0.30% rate per day, how many days will it take?

Answer 1

Hi Kiley,

This is an example of exponential decay (the starting amount is being reduced every day).

In general we have something like A = A₀(rate)^x. Where A is the number of units that remain after a given number of days "x" and A₀ represents the starting amount. Determining the rate of decay is the key for starting the problem. For this problem we have A = 2702 (.997)^x. Where did the .997 come from. First of all .3% is written as a decimal as .003, Every day .003 units disappear (decay). So what amount is left? Subtract .003 from 1.000 to get the rate of decay. In other words after 1 day 2702(.997) of the stuff remains.

Now how to we determine how many days it takes to reach 2316 units left? We need to solve this equation:

2316 = 2702(.997)^x. This can be solved by taking the natural log of both sides, after dividing both sides by 2702.

So, 2316/2702 = .997^x

ln(2316/2702) = ln .997^x

ln(2316/2702) = x ln.997 (using property of logarithms)

Finally x = ln(2316/2702)/ln .997

You could also use common log.Use your calculator to evaluate the expression to get the value for x (in days).