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Not sure how to explain this. Let's say we have a value that doubles every 12 years. By what percentage is it increasing each of those 12 years?

For example, if we start with 100, and it becomes 200 in 12 years, it increased 100% in 12 years. 100/12=8.33% per year. But increasing by 8.33% doesn't result in a 100% in 12 years:

Y1: 100 + 8.33% = 108.33 Y2: 108.33 + 8.33% = 117.35 Y3: 117.35 + 8.33% = 127.13 Y4: 127.13 + 8.33% = 137.72 Y5: 149.12 Y6: 161.62 Y7: 175.08 Y8: 189.67 Y9: 205.47

So by what percent does it actually need to increase each year to double in 12 years?

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  • $\begingroup$ The twelfth root of $2$ is $1.0594631$. So it increases by $5.94631$ percent each year. $\endgroup$
    – Phil H
    Sep 12, 2022 at 23:49

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Let's say we start with value $v$. Let $r$ be the fraction of the current value that we add each year.

Then after one year we have $v+rv=(1+r)v$.

The following year we have $(1+r)v+r(1+r)v=(1+r)^2v$

Continuing in this manner we get that $$(1+r)^{12}v=2v\implies (1+r)^{12}=2\implies r=\sqrt[12]{2}-1$$

Multiply that by $100\%$ to get the annual interest rate.

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It looks like considering an increase by $r = \frac{100}{12} \approx 8.33\%$, however that would only work if you were adding $8.33\%$ of the principal amount $\$100$, i.e. $\$8.33$ each year. What you are considering here involves compound interest, where the interest rate assessed against the amount after each year.

Let's call the initial amount $A_0$, then $A_1$ after one year, $A_2$ after two years, etc., all the way up to $A_{12}$ after $12$ years. In your example, $A_0 = 100$ and $A_{12} = 200$. Say that the unknown growth rate is called $r$. This is an actual number, not a percent, so e.g. $25\% = 0.25$.

Begin with $A_0$. Then, $A_1 = A_0 + rA_0$, representing the amount you already have plus the amount of growth. Notice that you can factor this expression to write $A_1 = (1+r) A_0$.

The same goes for $A_2 = A_1 + r A_1 = (1+r) A_1$. However, we can trace it all the way back to $A_0$, obtaining: $A_2 = (1+r)^2 A_0$. Similarly, $A_3 = (1+r)^3 A_0$, and so on.

Apparently, after $n$ years, beginning with principal amount $A_0$ and growing at rate $r$, we have $$ A_n = (1+r)^n A_0. \tag{1} $$

Now, you are interested in the situation where you know $n$, $A_0$, and $A_n$, and you're interested in $r$. In other words, you want to solve the equation $(1)$ for $r$: \begin{align} A_n &= (1+r)^n A_0 \\ \frac{A_n}{A_0} &= (1 + r)^n \\[2pt] \biggl( \frac{A_n}{A_0} \biggr)^{\! \frac{1}{n}} &= 1 + r \end{align} yielding the formula $$ r = \biggl( \frac{A_n}{A_0} \biggr)^{\! \frac{1}{n}} - 1 $$ Notice: the values of $A_0$ and $A_n$ don't matter here—just their ratio.

In your example, with $n=12$, $A_0 = 100$, and $A_{12}= 200$, you need to calculate $1$ less than the $12$th root of $2$: $$ r = \biggl( \frac{200}{100} \biggr)^{\! \frac{1}{12}} - 1 = 2^{\frac{1}{12}} - 1 \approx 0.0595 = 5.95\% $$


Incidentally, this number is relevant in music, as the Western chromatic equal-tempered scale (think keys on a piano) divides the octave into $12$ equally spaced intervals. What that actually means is the ratio of frequencies of consecutive notes is constant. Which constant? Well, the octave is always a frequency twice as big as the base tone, so again, we're calculating $12$th roots of $2$ to get the percent increase from one note to the next. You can listen to the graph here.

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    $\begingroup$ +1 for the comment about equal-tempered scale. Then as a quick mental calculation, (approximately) a perfect fifth above is $\frac32\times$ the original frequency or $7$ semitones above; the original value would become about $150$ after the first $7$ years. A perfect fourth above is $\frac43\times$ the original frequency or $5$ semitones above; the remaining $5$ years would grow the value by $\frac13$, from $150$ to $200$. $\endgroup$
    – peterwhy
    Sep 13, 2022 at 1:08
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If you don't know how to compute twelvth roots on a calculator or on google, just remember that $\sqrt[12]{x} = x^{(1/12)}$. Works with square roots, cube roots, and all the others too.

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