Estimating Logs and Powers of Two

Roughly how many times do you have to double $1 to pass $1,000,000? Questions like this are really asking for a log base 2, and most people freeze because "logarithm" sounds like it needs a calculator. It doesn't — powers of two and a couple of memorized log landmarks get you there in a few seconds.

Two small libraries to memorize

First, powers of two up through $2^{10} = 1{,}024$: $2,4,8,16,32,64,128,256,512,1{,}024$. Second, one crucial log10 landmark: $\log_{10}2 \approx 0.301$. Everything else follows from these two facts plus the change-of-base relationship:

$$\log_2 x = \frac{\log_{10} x}{\log_{10} 2} \approx \frac{\log_{10} x}{0.301} .$$

In plain English: to find "2 to what power gives x," find how many powers of 10 fit into x (that's $\log_{10}x$), then divide by $0.301$ since each power of 2 only advances you 0.301 of a power of 10.

Worked example 1: doubling $1 to $1,000,000

We want $\log_2(1{,}000{,}000)$. Since $\log_{10}(1{,}000{,}000) = 6$ exactly:

$$\log_2(1{,}000{,}000) \approx \frac{6}{0.301} \approx 19.9 .$$

So doubling roughly 20 times gets you from $1 past $1,000,000 — check it: $2^{20} = 1{,}048{,}576$, just over a million, confirming the estimate to within rounding. Notice how little arithmetic that took: no calculator, no long division, just one memorized constant and a single division.

Worked example 2: how many digits does 3^40 have?

Digit count relates to $\log_{10}$: a number has $\lfloor \log_{10}x \rfloor + 1$ digits. We need $\log_{10}(3^{40}) = 40 \log_{10}3$. Recall $\log_{10}3 \approx 0.477$ (worth memorizing alongside $\log_{10}2$). So:

$$40 \times 0.477 = 19.08 .$$

That means $3^{40} \approx 10^{19.08}$, which has $\lfloor 19.08 \rfloor + 1 = 20$ digits — a fact you'd never guess without logs, reached here in one multiplication and a floor. It's also a good example of why memorizing a couple of extra log10 landmarks beyond just $\log_{10}2$ pays off: $\log_{10}3 \approx 0.477$ and $\log_{10}7\approx0.845$ round out most of what you'll need for digit-count and order-of-magnitude questions.

What this means in practice

This shows up any time a problem involves compounding, doubling times, or "how big does this get" questions — estimating years to double an investment, gauging how many rounds of a binary search a dataset needs, or sanity-checking an order-of-magnitude Fermi estimate. The single fact $\log_{10}2 \approx 0.301$, combined with $2^{10}\approx 10^3$ as a memory anchor, unlocks nearly every log-based mental estimate you'll be asked for.