I try to understand the actual intuition behind the logarithm properties and came across a post on this site that explains the multiplication and thereby also the division properties very nicely:

Suppose you have a table of powers of 2, which looks like this: (after revision)

$$\begin{array}{rrrrrrrrrr} 0&1&2&3&4&5&6&7&8&9&10\\ 1&2&4&8&16&32&64&128&256&512&1024 \end{array}$$

Each column says how many twos you have to multiply to get the number in that column. For example, if you multiply 5 twos, you get $2\cdot2\cdot2\cdot2\cdot2=32$, which is the number in column 5.

Now suppose you want to multiply two numbers from the bottom row, say $16\cdot 64$. Well, the $16$ is the product of 4 twos, and the $64$ is the product of 6 twos, so when you multiply them together you get a product of 10 twos, which is $1024$.

I found that very helpful to understand the actual proofs for this property.

I still struggle to get the idea behind the change of base rule. I'm familiar with the proof that goes like:

$$\log_a x = y \implies a^y = x$$ $$\log_b a^y = \log_b x$$ $$y \cdot \log_b a = \log_b x$$ $$y = \frac{\log_b x}{\log_b a}$$

But can somehow provide a explanation in the style of the quoted answer why this actually works?

  • $\begingroup$ Small nit-pick, $a$ must be positive, otherwise you can get the false statement $\log_b((-1)^2)=2\log_b(-1)$. $\endgroup$ – Simply Beautiful Art Dec 28 '17 at 2:38
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    $\begingroup$ One idea: use the same table to rationalize why $\log_4(x) = \frac{\log_2(x) }{ \log_2(4) } = \frac{1}{2} \log_2(x)\,$. $\endgroup$ – dxiv Dec 28 '17 at 2:42
  • $\begingroup$ One interesting thing about reading all these answers is that I'd always thought of the change of base formula in the divided form, $ \log_a b = \frac{\log_c b}{\log_c a}$, and hadn't tried to develop any intuition behind it, just applying it rotely. But now the multiplicative form, $ \log_a x = \log_a b \times \log_b x$, has a lot more meaning to me. Of course the two are mathematically equivalent, but it's weird how intuition can latch onto one more easily than the other. $\endgroup$ – JonathanZ Dec 28 '17 at 18:19
  • $\begingroup$ Many excellent answers, I also didn't expect that the question gets this much attention. Hard to pick any answer specifically, as many of them contributed to my understanding. $\endgroup$ – Max Dec 29 '17 at 4:01
  • $\begingroup$ @SimplyBeautifulArt: And $b = 1$ disproves Max's claim that it actually works. $\endgroup$ – user21820 Dec 29 '17 at 6:43

Here is one way of looking at it. (I'll assume that the numbers $a,b,x \in \mathbb R$ satisfy $a > 1, b > 1$, and $x > 0$.)

I dislike the name "logarithm" and I think a more descriptive name for $\log_b(x)$ is "the exponent from $b$ to $x$". We could also use the notation $[b \to x]$ instead of $\log_b(x)$. The change of base rule then tells us that the exponent from $b$ to $x$ is equal to the exponent from $b$ to $a$ times the exponent from $a$ to $x$: $$ \tag{$\spadesuit$}[b \to x] = [b \to a][a \to x] $$ or equivalently $$ [a \to x] = [b \to x]/[b \to a]. $$ In standard notation, this formula states that $$ \log_a(x) = \frac{\log_b(x)}{\log_b(a)}. $$

Note that equation $(\spadesuit)$ is obvious, because \begin{align} b^{[b \to a][a \to x]} &=(b^{[b\to a]})^{[a \to x]} \\ &= a^{[a \to x]} \\ &= x. \end{align}

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    $\begingroup$ I agree with your first sentence. Sometime it helps me to read 'logarithm' as 'what's the exponent?'. $\endgroup$ – JonathanZ Dec 28 '17 at 3:25
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    $\begingroup$ Please be more precise, otherwise your claims are false. Try $b=1$. $\endgroup$ – user21820 Dec 28 '17 at 11:31
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    $\begingroup$ @user21820 I probably should have stated that $a$ and $b$ are greater than $1$ and $x$ is positive, but the question was only asking for intuition and I wanted to get to the point as fast as possible. $\endgroup$ – littleO Dec 28 '17 at 11:45
  • $\begingroup$ @user21820 Just updated the answer as you suggested. $\endgroup$ – littleO Dec 29 '17 at 20:46

Intuition is always tricky to get across, but I can try.

$\log_bx$, as you noted, tells you how many $b$s you need to multiply together to get $x$. Now if you need $\log_ba$ number of $b$s to multiply to get $a$, and you need $\log_ax$ number of $a$s to multiply to get $x$, we can "expand" each of those $a$s into a number of $b$s. There will be $\log_ba$ number of $b$s for each $a$, so the total number of $b$s will be $\log_ax \log_ba$. These $b$s multiply to $x$, so $\log_ax \log_ba = \log_bx$.

For example, take $b=2, a=8, x=64$. We start with $\log_ax = 2$, which tell us we need two $8$s to get $64$:

$$ 8 \cdot 8 = 64 $$

We use $\log_ba = 3$, i.e., $2 \cdot 2 \cdot 2 = 8$, to expand each $8$:

$$ (2 \cdot 2 \cdot 2) \cdot (2 \cdot 2\cdot 2) = 64 $$

Now the total number of $2$s we are multiplying is $2 \cdot 3 = 6$, so $log_2 64 = 6$


It might help to crank though some easy examples

$4096 = 2^{12} = (2^2)^6 = (2^3)^{4}$

$\log_2 4096 = 12\\ \log_4 4096 = \frac {12}{2} = \frac {\log_2 4096}{\log_2 4}\\ \log_8 4096 = \frac {12}{3} = \frac {\log_2 4096}{\log_2 8}\\ \log_{16} 4096 = \frac {\log_2 4096}{\log_2 16} = 3 \implies (2^4)^3 = 2^{12}$


Motivated by @dxiv's comment, we can adapt your table to understand the change of base rule, although I'm just going to illustrate changing base from $2$ to $4$ and prove the formula in the (equivalent) form $$ \log_2 x = log_2 4 \times \log_4 x. $$ Let's add a row to your table:

$$\begin{array}{rrrrrrrrrr} \text{exponent}& 0& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ \text{power of 2}&1& 2& 4& 8& 16& 32& 64& 128& 256& 512& 1024\\ \text{power of 4}&1& 4& 16& 64& 256& 1024& 4096& 16384& 65536& 262144& 1048576 \end{array}$$

The top row, labeled exponents, is where you read off the logarithms, and they can be to the base $2$ or the base $4$ depending on which lower row you are looking at.

Imagine that there are counters in the bottom two rows, and they both start in the first column, on their $1$. Start moving the bottom counter forward, one column at a time. Now suppose the 'powers of 2' counter starts moving forward and it tries to stay on the same number as the 'powers of 4' counter is on. So when 'powers of 4' jumps to '4', 'powers of 2' has to move two columns to get to its 4. When 'powers of 4' then moves to 16, 'powers of 2' again has to move two columns to get to its 16. I.e. the powers of 2 counter has to move twice as fast as the powers of 4 counter. That means when you read off the logarithms from the top row $$\log_2 x = 2 \times \log_4x.$$

If you added a 'powers of 8' row you could easily see that $\log_2 = 3 \times \log_8x$, and then generalize it to compare $\log_a$ and $\log_{a^n}$.

  • $\begingroup$ This is a wonderfully intuitive answer that used the OP's framework to address the question. Thank you and +1 from me! $\endgroup$ – Slecker Oct 20 '18 at 21:47
  • $\begingroup$ Thank you. I really liked the dynamic image of $\log_2$ traveling faster than the $\log_4$. Glad you did too. $\endgroup$ – JonathanZ Oct 20 '18 at 23:48

Will it help to write the formula in a different way:

$$\log_a c=\log_a b\times\log_b c $$


This is motivated as follows: assume you start with $a $ and you want to "reach" $c $ by exponentiation.

  • You can first raise $a $ to some exponent $x=\log_a b$ to get $b=a^x$, and then you can raise $b $ to some exponent $y=\log_b c$ to get $c=b^y=(a^x)^y=a^{xy}$...

  • ... Or, you can directly raise $a $ to the exponent, well, $xy=\log_a b\times\log_b c$ to get $c $. The thing is, we call this last exponent $\log_a c $.

Thus, $\log_a c=\log_a b\times\log_b c $.


Take a slide rule. There is a linear scale on it (on this photo it's labeled as lgX). It is deliberately created to be linear with respect to length because this allows multiplying numbers by adding corresponding line segments. This ability is the main advantage of a slide rule.

Now imagine there are two such scales. In the example below they are the first two:

0-------1-------2 $\log_4 x$
0---1---2---3---4 $\log_2 x$
1---2---4---8--16 $x$

The trick is: whatever (sane) bases you take, if the first scale is linear, so is the second. More general picture:

0----1------p---... $\log_a x$ (linear)
0----q------1---... $\log_b x$ (linear)
1----a------b---... $x$ (non-linear)

The two "upper" scales are linear, they have $0$ at the same place, so there is a simple proportion behind them. For every (sane) $x$ we have:

$$\frac {\log_a x} {\log_b x} = \frac 1 q = \frac p 1$$

where obviously $p=\log_a b$ and $q=\log_b a$.


I understand the change of base formula as a graph transformation hinted by the comment by dxiv. The change of base formula encapsulates the idea that all logarithms are equivalent up to vertical dilations (and reflections). More precisely, if you have two positive numbers $a$ and $b$, then there is a constant $C$ such that $$\log_a(x)=C\cdot\log_b(x)$$ for every positive $x$. The change of base formula tells you that $C=1/\log_b(a)$.

This can be paralleled with exponentials in a similar way, for there is a "change of base formula" for exponentials. But it doesn't get the same limelight as it does for logarithms. All exponentials are equivalent up to horizontal dilations (and reflections). More precisely, if $a$ and $b$ are positive numbers again, then there is a constant $C$ such that $$a^x=b^{C\cdot x}$$ for every real $x$. This time $C=\log_b(a)$. Interestingly, the $C$'s in this discussion are inverses of each other (which shouldn't be a surprise, but I never noticed before).


Similar to @JonathanZ's table, try this

\begin{array}{c|rrrrrrrrr} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2^n & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ 8^{(n/3)} & 1 & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 & 512 & 1024 \\ \end{array}

Hence $\log_8(x)= \dfrac{\log_2(x)}{3} = \dfrac{\log_2(x)}{\log_2(8)} $


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