# Find first power of 2 with exponent that is a multiple of 10 whose first number is not 1

I've been pondering this quirky fact about powers of 2 for a while now, and I can't seem to formulate it properly in my head to find a proper answer without just using a calculator and trying out numbers till I find the answer.

When talking about memory we usually use Megs, Gigs, etc. And usually these are technically powers of 2.

Kilo = pow(2,10) Meg = pow(2,20) Gig = pow(2,30)

Ok so it seems that 2 to the power of something divisible by 10 is going to be the approximation of 10 to the power of something divisible by 3 (which seems coincidental even strange to me), and by approximation I mean the first digit is a 1 and it has the correct number of digits.

Using experimentation it appears pow(2,299) is the first time this doesn't hold true, but I can't help but think there must be a way to make a formula that gives me the value without experimentation.

I know for example that log 2 is .30102... which is the reason it works for sufficiently small exponents since we want 3 extra digits each time. But what's next?

• Feb 24 '13 at 3:17
• Just to follow up on this, pow(2,299) is the best approximation for pow(1000,30) or 30 orders of magnitude, where "Gig" is 3 orders. I don't think 30 orders of magnitude has an official name. I think it should have one, if it doesn't already. Just so that I can give this phenomenon a name. Jan 18 '16 at 22:50
• @ZevChonoles you are correct though I was referring to finding the power of 2 that approximates the orders of magnitude which are powers of 1000, and in this case we see that not all of the exponents of these approximations are divisible by 10 Jan 18 '16 at 22:56

It sounds like you're asking the first value of $n$ such that

$$10^{3n} < 2^{10n} < 2 \cdot 10^{3n}$$

is false. It will be false because the second inequality is violated, so you seek

$$2^{10n} > 2 \cdot 10^{3n}$$ $$2^{10n-1} > 10^{3n}$$ $$(10n-1) \log 2 > 3n \log 10$$ $$n(10 \log 2 - 3 \log 10) > \log 2$$ $$n > \frac{\log 2}{10 \log 2 - 3 \log 10}$$ $$n > 29.226\ldots$$

(The inequality doesn't reverse when I divide, because I divide by a positive number).

• This seems right to me. The first integer n where it is false is 30 or pow(2,300) so n > 29.2 makes sense. Thanks for the nice equations. Feb 24 '13 at 19:42

So $\log_{10}2^{10k}=(3.010299\dots)k=3k+(.010299\dots)k$, and we want the smallest $k$ such that $(.010299\dots)k\gt\log_{10}2$. So $$k={\log_{10}2\over10\log_{10}2-3}$$ which is a bit under $30$, and the power we want, $10k$, is thus a bit under $300$.

• It seems you have the right equation but I don't follow how you arrived there. I voted it up but the other answer with the same end result explained it a little better, so I accepted that one. Feb 24 '13 at 19:45

As $$2^{10n} = (2^{10})^n = 1024^n = (1.024\times 10^3)^n = 1.024^n\times 10^{3n},$$ you want to find is the smallest $n$ such that $1.024^n \geq 2$. Taking the natural logarithm of both sides, and noting that $\ln x$ is an increasing function on $(0, \infty)$, we have

\begin{align*} \ln 1.024^n &\geq \ln 2\\ n\ln 1.024 &\geq \ln 2\\ n &\geq \frac{\ln 2}{\ln 1.024} \approx 29.23 \end{align*}

so $n = 30$ is the smallest such $n$; as $\ln x$ is a strictly increasing function on $(0, \infty)$ and $1.024 > 1$, $\ln 1.024 > \ln 1 = 0$, so dividing both sides of the second line does not change the inequality.

Therefore $2^{10\times 30} = 2^{300}$ is the first power of $2$, with exponent divisible by $10$, which does not begin with a $1$.