Introduction
Working with huge numbers, we often take a logarithm to "shrink" them. A trillion (\(10^{12}\)) has 13 digits when written out, but taking its logarithm collapses it down to a short, plain 12. But here's something odd — some textbooks write logarithms with base 10, others with base 2, and others use the natural logarithm (base \(e\)). Does it really not matter which base you pick?
This post ties two questions together. ① Why does a logarithm "shrink" big numbers the way it does? ② Why is it fine to switch the base? Both answers fall out of the same picture: a logarithm is a ruler with evenly spaced marks.
1. A logarithm is a ruler that counts "×10 steps"
Let's grow a number ×10 at a time and watch how its logarithm moves alongside it.
| \(N\) | digit count | \(\log_{10}N\) |
|---|---|---|
| 1 | 1 | 0 |
| 10 | 2 | 1 |
| 100 | 3 | 2 |
| 1,000 | 4 | 3 |
| 10,000 | 5 | 4 |
\(N\) multiplies by exactly 10 every step, exploding all the way up to ten thousand in no time — while \(\log_{10}N\) on the right climbs by a calm, steady 1 every single step. Let's confirm this isn't a coincidence, starting from the definition.
The statement \(\log_a M = p\) means exactly "raising \(a\) to the \(p\)-th power gives \(M\)" — in other words, it's just another way of writing \(a^p = M\). Here \(a\) is called the base (10, in the table above), and \(M\) is the number the logarithm is being taken of.
Plugging this definition straight into the table checks out: since \(N=10^0=1\), \(\log_{10}1=0\); since \(N=10^1=10\), \(\log_{10}10=1\); and so on, matching every row exactly. This relationship isn't limited to \(N\) landing exactly on a power of 10, either — it flows smoothly through the values in between. Take \(N=999\): it's still a 3-digit number, and \(\log_{10}999\approx2.9996\) sits just barely under 3. The instant \(N\) becomes \(1{,}000\), \(\log_{10}1000=3\) crosses over to a whole integer, and the digit count crosses over to 4 digits at exactly the same moment. In other words, the digit count of the integer part of \(N\) is always exactly 1 more than the integer part of \(\log_{10}N\).
The reason the logarithm climbs this calmly is that each mark on this ruler doesn't measure "how much was added" — it measures "how many times bigger." Every time \(N\) multiplies by 10, the logarithm's ruler advances by exactly one mark (1). That's why 100 million (\(10^8\)) and 1 (\(10^0\)) differ by a factor of a hundred million in actual value, but only by 8 in logarithm value — a logarithm is exactly this: compressing multiplicative explosions down into additive, gentle change.
2. Why you can change the base — it's just a choice of ruler unit
Every logarithm carries a base. Writing \(\log_{10}\) means the base is 10; writing \(\log_2\) means the base is 2. Using \(e\,(\approx2.718)\) as the base gets its own name, the natural logarithm, written \(\ln\). Measure the same number with a different base and you get a different logarithm value — exactly like measuring the same distance in centimeters versus inches gives you different numbers on the ruler.
So what relationship holds between logarithms taken in different bases? Let \(x=\log_a M\). By the definition we just confirmed, this means:
$$a^x = M$$Now let's take the base-\(b\) logarithm of both sides.
$$\log_b(a^x) = \log_b M$$The left side is just the power rule \(\log_b(a^n) = n\log_b a\) derived in Why Do Logarithms Turn Multiplication Into Addition? — swap in \(n=x\) and it applies directly.
$$x\log_b a = \log_b M$$Dividing both sides by \(\log_b a\) isolates \(x\).
$$x = \frac{\log_b M}{\log_b a}$$But \(x\) was originally \(\log_a M\), so we can write:
$$\log_a M = \frac{\log_b M}{\log_b a}$$This is the change-of-base formula. When you want "the logarithm measured in base \(a\)," but your calculator only has buttons for the common logarithm (base 10) or the natural logarithm (base \(e\)), this formula lets you get there anyway. For instance, to compute \(\log_2 8\) using a calculator's common logarithm:
$$\log_2 8 = \frac{\log_{10}8}{\log_{10}2} \approx \frac{0.903}{0.301} = 3.000$$Checking directly, \(2^3=8\), so this matches exactly.
3. So why does the ratio never change — same ratio, different units
Whether the distance from Seoul to Busan is how many times farther than the distance from Seoul to Daejeon comes out to the same multiple whether you measure both in kilometers or in miles. Changing units scales both distances by the same factor, so the "how many times" relationship between them survives untouched. A logarithm's base plays exactly the same role.
Consider two numbers \(M_1,\ M_2\), and look at the ratio of their logarithms in base \(a\). Applying the change-of-base formula we just derived to each one separately:
$$\frac{\log_a M_1}{\log_a M_2} = \frac{\dfrac{\log_b M_1}{\log_b a}}{\dfrac{\log_b M_2}{\log_b a}}$$The \(\log_b a\) appearing identically in both the numerator and denominator cancels out completely in the division.
$$\frac{\log_a M_1}{\log_a M_2} = \frac{\log_b M_1}{\log_b a} \times \frac{\log_b a}{\log_b M_2} = \frac{\log_b M_1}{\log_b M_2}$$The left side (the ratio measured in base \(a\)) and the right side (the ratio measured in base \(b\)) are exactly equal. Notice that no particular \(a\) or \(b\) survives anywhere in this result — so whether you measure with base 2, base 10, or base \(e\), the ratio between two numbers' logarithms is always the same.
Checking with actual numbers: for \(M_1=16,\ M_2=4\) (since \(16=4^2\), the answer should come out to 2):
$$\frac{\log_2 16}{\log_2 4} = \frac{4}{2} = 2, \qquad \frac{\log_{10}16}{\log_{10}4} \approx \frac{1.204}{0.602} = 2.000, \qquad \frac{\ln 16}{\ln 4} \approx \frac{2.773}{1.386} = 2.000$$All three bases give exactly 2. A logarithm's base is nothing more than choosing a unit for the ruler — whichever unit you choose, the "how many times" relationship between two numbers stays fixed.
4. Try it yourself
The top interactive covers Section 1. Drag the slider for the exponent \(t\), and you'll see just how explosively \(N=10^t\) grows on one bar, while the logarithm \(\log_{10}N=t\) fills in steadily and evenly across the same range on the ruler beside it. Watch how the integer-part digit count ticks up by exactly one every time \(t\) crosses a whole number.
The bottom interactive covers Sections 2 and 3. Pick two numbers \(M_1,\ M_2\) with the sliders, and switch between base 2, 10, and \(e\) with the buttons to see how each logarithm value (bar length) changes with the base — and confirm that the ratio between them stays exactly the same no matter which base you pick.
Key takeaways
- When \(N\) multiplies by 10, \(\log_{10}N\) climbs by exactly 1 — a logarithm is a ruler that counts "×10 steps."
- The digit count of the integer part of \(N\) is always exactly 1 more than the integer part of \(\log_{10}N\) (e.g., \(N=1{,}000\) → \(\log=3\) → 4 digits).
- The change-of-base formula \(\log_a M = \dfrac{\log_b M}{\log_b a}\) falls directly out of the logarithm's power rule, \(\log_b(a^n)=n\log_b a\).
- The ratio between two logarithms, \(\dfrac{\log_a M_1}{\log_a M_2}\), doesn't depend on the base \(a\) at all — the \(\log_b a\) term cancels out completely in the computation.
- A logarithm's base is just a choice of ruler unit — change the unit, and the "how many times" relationship is preserved.
A logarithm isn't some magic trick that shrinks big numbers — it's simply a ruler that rewrites the world of multiplication in the language of addition. Whatever unit (base) you pick for that ruler, the ratio between two numbers measured on it never changes.
Related reading — Why Do Logarithms Turn Multiplication Into Addition? (where the power rule used in this post comes from) · Why Is a⁰ Always 1, and Why Does a Half Power Become a Square Root? (extending exponents, the mirror direction of a logarithm)