# Ask Uncle Colin: A logarithmic coincidence?

Dear Uncle Colin,

I noticed that $2^{\frac{1}{1,000,000}} = 1.000 000 693 147 2$ or so, pretty much exactly $\left(1 + \frac{1}{1,000,000} \ln(2)\right)$. Is that a coincidence?

Nice Interesting Numbers; Jarring Acronym

Dear NINJA,

The easiest way to see that it’s not a coincidence is to check out $3^{\frac{1}{1,000,000}} $, which is about $\left(1 + \frac{1}{1,000,000} \ln(3)\right)$ – but that leaves unanswered the question “why?”.

My first thought was to use the binomial theorem – to check that $\left(1 + \frac{1}{1,000,000} \ln(2)\right)^{1,000,000} \approx 2$, or in general, $\left(1 + \frac 1n \ln(k)\right)^n \approx k$, you’d expand it to get $1 + \ln(k) + \frac{1}{2}(n-1) (\ln(k))^2 + …$, which actually isn’t all that helpful ((That said, it does give an interesting identity involving powers of a natural logarithm.)).

Instead, the trick is to use logarithms straight off, and rewrite the left-hand side as $e^{\ln\left(2^{\frac{1}{1,000,000}}\right) } = e^{\frac {1}{1,000,000} \ln(2)}$. For small $x$, $e^x \approx 1 + x$, so $e^{\frac {1}{1,000,000} \ln(2)} \approx 1 + \frac{1}{1,000,000} \ln(2)$. This, clearly, works if you replace 2 or 1,000,000 with whatever numbers you like.

-- Uncle Colin