# At 3% inflation rate, is \$100 today worth \$40 20 years ago?

I'm little bit confused... At 3% inflation rate, is \$40 20 years ago worth \$100 today? Is it as simple as adding 60% of \$100 to get the worth of currency 20 years ago?

Forgot about compound interest... It makes sense now. \$40 20 years ago would be worth \$72,24 today at 3% inflation.

• \$40 31 years ago would be worth exactly \$100 today Aug 3, 2022 at 12:27
• How much worth would It be 40 years ago? According to your calculation you'd subtract 40 * 3 = 120 percent, so something must be wrong. Aug 3, 2022 at 18:23
• No one's mentioned the Rule of 72, so I will: investopedia.com/terms/r/ruleof72.asp -- at 3%, money halves in value every 24 years, so decreasing to 40% in 20 years can't be accurate Aug 5, 2022 at 13:23

Inflation diminishes value, so the older value would be greater, e.g.

At 3% inflation, 20 years ago \$180.61 is today worth \$100

``````100 (1 + 0.03)^20 = 180.611
``````

and in 20 years \$100 today will be worth \$55.37

``````100/(1 + 0.03)^20 = 55.3676
``````

\$55.37 at 3% interest for 20 years would be \$100 today

• This is not how changes in the value of a dollar is usually presented. We usually say that if someone won \$40 20 years ago, it's equivalent to winning \$100 today, because they both can buy the same amount of stuff in their respective time periods. Similarly, a job that paid \$40K/year 20 years ago would pay \$100K today. Aug 3, 2022 at 19:43
• Math nitpick: The math here is backwards. Although the value of the dollar goes down, the numbers should go up. \$180 worth of buying power in 2000 will require more money in 2020, because each dollar is worth less. The statements made are only correct if you measure today-dollars in 20-years-ago dollar values, but a much less confusing way of saying that is "\$100 in 2000 dollars is equivalent to \$180.61 in 2020 dollars" Aug 4, 2022 at 4:46
• @Barmar A job that paid \$40K/year 20 years ago would pay \$100K today — uh-uh, only if salaries keep up with inflation, which is not at all a given. Aug 4, 2022 at 6:43
• This answer is simply incorrect. Aug 4, 2022 at 19:29
• the answer is correct, just not how it's usually presented. If I acquire a \$100 bill today and hold onto it for 20 years during 3% inflation, it will then have the equivalent purchasing power that \$55 and change has today. On the other hand, if I spend that \$100 bill today on goods that track inflation, then sell them in 20 years, I will receive \$180 Aug 4, 2022 at 22:43

\$100 20 years ago is still worth \$100 today. What changes is what that \$100 can buy. Inflation is a measure of how purchasing power changes, and compounds over time.

So a better explanation would be: What cost `X` today cost `Y` 20 years ago. That calculation would be:

``````100 / (1 + 0.03)^20 = 55.36
``````

or, you could say what costs \$55.36 20 years ago costs \$100 today.

But note that it is a very broad measure and does not necessarily apply to any one product, or even cost of living in general. Many products vary in price differently than actual inflation and can swing up and down significantly (like gasoline). Cost of living can also change differently depending on where you live. I don't have data, but I would suspect that the cost of living in San Francisco has risen much higher than "inflation" over the last 20 years, and some areas have increased less than "inflation".

• When you say, "\$100 20 years ago is still worth \$100 today," it would probably be helpful to specify that you're talking about the nominal value, not the real value. The real value of \$100 in 2002 dollars is definitely not equal to the real value of \$100 in 2022 dollars. It's true enough that if you stuck a \$100 bill under your mattress in 2002, it would still be nominally worth \$100 in 2022, but its real value will have decreased significantly in that time. The real value of \$100 is literally defined in terms of what it can buy. Aug 4, 2022 at 8:37
• Just wondering, why do you do (1 + 0.03) and not just 1.03? I guess it has to do with some correctness, but whats the logic behind it? 0.03 is just the percentage, but why not write it like 1.03? Aug 5, 2022 at 7:49
• @Allart Just to illustrate that it's 1 + the percentage - formulas are typically written in terms of `1+p` or `1+r`. Aug 5, 2022 at 15:03

Inflation means that the prices of goods and services are increasing. \$40 is still \$40 20 years later. However, because prices have increased 3% per year in your example, you cannot buy as many goods and services with \$40 today as you could 20 years ago.

At 3% inflation rate is \$40 20 years ago worth \$100 today?

At a 3% inflation rate, what you could buy 20 years ago with \$40 would cost \$72.24 today.

It's not as simple as adding 60% of \$100 to get the worth of currency 20 years ago?

As you pointed out in your edit, you cannot simply multiply the inflation rate by the time period. If inflation is 3% per year, prices increase 3% over the prior year every year. Taking 3% * 20 would be as if prices increased 3% of their initial value each year.

The formula for calculating prices 20 years from now at 3% inflation per year would be the same as the formula for calculating compound interest:

``````\$40 (1 + 3%/100)^(20 years) = \$72.24
``````

The formula for working backwards to calculate prices 20 years ago at 3% inflation per year for goods worth \$100 today would be:

``````\$100.00 / (1 + 3%/100)^(20 years) = \$55.37
``````
• The `%` already includes the `/100` in its meaning, so writing `%3/100` means you are out by a factor of 100. Either you write `(1 + 3%)^(20 years)` or you write `(1+3/100)^(20 years)`. Aug 3, 2022 at 12:42

Your math is way off. I mean you got the intuition right that inflation means that money loses it's value, so a small number in the past is equivalent to a large number today.

But if you'd just add 60% to \$40 then the result would be \$64:

\$40 + \$40 *60%

\$40 + \$40 * 60/100 (percent = per centum = per hundred)

\$40 * (1 + 60/100) = \$40*(1+0.6) = \$40*1.6= \$64

Now that's too low, so either you're estimate of \$40 is wrong and/or you're inflation rate is wrong (too low and or computed wrongly).

So if the cumulative inflation over the last 20 years would indeed be 60% then you'd need to solve the equation:

X *1.6 = 100

or conversely:

X = \$100/1.6 = \$100 / (160/100) = \$100 * 100 / 160 = \$62.5

However there is another problem in your calculation and that is that the assumption of a cumulative inflation of 60% is wrong because the 3% increase are applied every year so you have spill over effects. To drive home that point lets consider an inflation rate of 100% per year. So each year everything gets twice as expensive. Now with your naive assumption of a linear application of the inflation that would be 2000% or 20 times as expensive. However as it's applied each year, you'd not have to double the initial amount, but the amount of last year. So if you start with \$1 twenty years ago, the next year that would correspond to \$2, then to \$4 the year after then to \$8 then \$16 and at year 5 you'd already exceed the x20 multiplier as it went to \$32. After 20 year it would be at 2²⁰ or `2¹⁰*2¹⁰ = 1024*1024 = \$1.048.576` (and yes I mean more than a million rather than \$20!!!) so that the x20 multiplier seem negligible to the actual value.

So you have a kind of snowball effect or exponential growth. Now luckily the inflation is not 100% but just 3%, but this still means it's not 60% increase over 20 years but (1.03)²⁰ = 80.6% or if you started from your \$40 it would be ~\$72 instead of \$64. Which is 20% more compared to the original \$40 (20% of that = \$8) and it's a 33% increase of the cumulated inflation compared to your estimate of 60% (60 * 1.33 = 80).

Which is still below 100 so the new number would be :

X *1.8 = 100

X = 100/1.8 = \$55.56

Now last but not least what would be the inflation rate if it had been \$40 twenty years ago and is \$100 now? Well that would be:

40 * (1+X/100)²⁰ = 100

Now you divide both sides by 40:

(1+X/100)²⁰ = 100/40 = 5/2 = 2.5

Then you apply the logarithm to each side:

log((1+X/100)²⁰) = log(2.5)

And as for logarithms that exponential 20 is the same as multiplying with 20:

20*log(1+X/100) = log(2.5)

log(1+x/100) = log(2.5)/20

Now let's say you picked ln for the log than that would be:

ln(1+x/100) = 0.04581

apply exp to both sides (log and exp are not perfect equivalence operations so be careful here; but for our purposes it should work)

exp(ln(1+x/100) = exp(0.04581)

exp and ln cancel out:

1+x/100 = 1.04688

x = (1.04688 -1)*100 = 4.688

So you'd have an inflation rate of 4.688% per year. And while you should check out how logarithms work you could also just estimate it by checking 1.03²⁰ = 1.8 which is smaller than the expected 2.5, 1.04²⁰ = 2.191 which is still smaller than the 2.5 and for 1.05²⁰ = 2.65 it's already above 2.5 so it's somewhere between 4% and 5% closer to 5% and then you further pick numbers between that and see if the value is higher or lower than what you expect and continue until you're satisfied with the accuracy.

So yeah that kind of math is very important to keep in mind because 5% return on investment per year yields more than 100% over 20 years but even more importantly if you loan at 20% that's WAY more than paying 4 times what you loaned over 20 years (closer to 40 times). So be aware of those pitfalls.