Wikipedia:Reference desk/Archives/Mathematics/2023 July 1

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July 1

Inflationary exponential growth

An item costs $440 on 1 July 2023. Assume in this scenario inflation is expected to run at 20% per annum forever. On 1 July 2024/25/26 the item is expected to cost $480, $576, $691.2 and so on.... But this only tells us what the price will be each July 1st. I want to know by which date the item will cost $600. Using round about maths I can roughly estimate that in 622 days it will reach $600 (using: 440*1.2^(622/365.3)), but the 622 days is my own trial and error input. There must be a more precise way of working this out which would give me an accurate decimal of 622 days. The number of days per year in this scenario should be treated as 365.3 rather than 365. Thanks Uhooep (talk) 12:59, 1 July 2023 (UTC)

The logarithm of daily inflation is $\log(1.2)/365.3$. You need to divide $\log({\frac {600}{440}})$ by that number; equivalently, your day number is ${\frac {365.3\log(15/11)}{\log(6/5)}}$. I get 621.4. By the way, why 365.3? —Tamfang (talk) 16:33, 1 July 2023 (UTC)

Length of a tropical year probably, although 365.3 is also probably a misrounding from 365.25, when the real value is closer to 365.24. GalacticShoe (talk) 18:22, 1 July 2023 (UTC)

If a quantity grows exponentially at a fixed rate, it should always increase by the same factor for a fixed period of time. Yet

•    480 / 440   =  1.09091
•    576 / 480   =  1.20000
•  691.2 / 576  =  1.20000

So I think "440" in the question should have been "400". Using that value, and the average length of a year in the Gregorian calendar (365.2425 days), the elapsed time comes out at 812.26 days.  --Lambiam 22:24, 1 July 2023 (UTC)

Good catch! --Stephan Schulz (talk) 11:51, 7 July 2023 (UTC)