Phrased slightly differently, for a fixed interest rate, growing from $500 to $1,000 takes the same amount of time as growing from $5,000,000 to $10,000,000.
Cool plots
The first plot assumes an initial investment on the x-axis, and shows the years required for it to grow by $100,000 on the y-axis:
It is just so much faster with $1,000,000 than with $100,000.
It is just so much faster with $1,000,000 than with $100,000.
The second plot shows the value of an initial $100,000 investment over time, and shades the periods required for $100,000 gains to show how much smaller it is later in the growth:
How does this all work?
Math behind this
The first problem said it took 40 years to go from $100,000 to $1,000,000 and wants to know how long it will take to go from $1,000,000 to $2,000,000. That is, if it takes 40 years to increase by a factor of 10, how long does it take to increase by a factor of 2? The equation we'll need is:
$$(1 + rate)^{time} = gain$$
First, we need the interest rate that yields a 10x increase in 40 years:
$$(1 + rate)^{40} = 10$$ $$40*log(1 + rate) = log(10)$$ $$log(1 + rate) = \frac{log(10)}{40}$$ $$rate = 10^{\frac{log(10)}{40}} - 1$$ $$rate \approx 0.059 $$
So the interest rate is roughly 5.9%. Now we just see how long it takes to double with that interest rate:
$$(1.059)^{t} = 2$$ $$t*log(1.059) = log(2)$$ $$t \approx 12$$
Since our time is in years here, that means that it takes approximately 12 years to double if it takes 40 years to increase by 10x.
From basically the same math as above, there's a simple relationship between increases:
$$\frac{t2}{t1} = \frac{log(gain 2)}{log(gain 1}$$
Testing it out, for gains of 10 and 2, we get that t2/t1 is ~3.3, and 40 / 12 ~ 3.3. Thus, a much faster way to solve that original problem is just:
$$\frac{40}{t} = \frac{log(10)}{log(2)}$$ $$t = 40*\frac{log(2)}{log(10)} $$ $$ t \approx 12 $$
And that's it...compound interest makes money grow really quickly (debt also). A bunch of idioms reference this, but it's cool that it's so easy to demonstrate.
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