What exactly is the base of a logarithm ? and how should it be understood ? I used to think it was the base of a "normal" exponent e.g. the $2$ in $2^{75}$ would be the base in logarithmic form, but the change of base formula can accept ANY base, and when finding the number of digits in $2^{75}$, you use the common log:




$$75(0.301)+1=23 \textrm{ digits}$$

I understand that these formulas work, I just can't wrap my head around why they work the way they do, and the heart of my issue is how I should understand the base.

I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that. But if that were the case, then would a base 16 represent a hexadecimal system ? and how would that work considering we use letters in addition to numbers ?

  • $\begingroup$ How many digits in $10^k$? and what is value of $\log_{10}(10^{k})$ ? Try searching for pattern. $\endgroup$
    – King Tut
    Feb 1 '18 at 17:45
  • $\begingroup$ "I did ponder that maybe a base of 10 represents a decimal system, and a base 2 would represent a binary system, but I haven't found any validation for that." Really? I would have thought those were the definitions. $\endgroup$
    – fleablood
    Feb 1 '18 at 18:00

I think you have the idea.

If $y = 2^{75}$ then $\log_2 y = 75$

One way to use bases is to find the number of digits that that number would have in that system. If $32<y<64$ then $5<\log_2 y < 6$

You can take logarithms in any base, and convert between bases.

$\log_a x = \frac {\log_b x}{\log_b a}$

For example, $\log_{10} 2^{75} = 75 \log_{10} 2$ as you have above.

But you could also say $\log_{10} 2^{75} = \frac {\log_2 2^{75}}{\log_2 10}$

Which implies $\log_2 10 = \frac 1{\log_{10} 2}$


You're right about the number of digits of a number in a specific base but why? To prove it let $(x)_b$ denote the number x in base $b$. If $b=10$ it is the same natural and convenient base we often use. Obviously if any number is between two consecutive powers of $10$, for example $1253$ is between $10^3$ and $10^4$ and it has 4 digits. In fact for any number $x$ if it belongs to $(10^n,10^{n+1}-1)$ for some n then $n+1$ is the number of digits of $x$. Then we can find the number of digits as following:$$10^n\le x<10^{n+1}\\n\le \log_{10}x<n+1\\n=\lfloor \log_{10}x\rfloor+1$$the same argument is true for any arbitrary base $b$, so the number of digits in base $b$ is $$dig_{b}(x)=\lfloor \log_{b}x\rfloor+1$$


The base of a logarithm is the base of the corresponding exponential function. Take $10^x$ for example. The base of this exponential function is $10$. The corresponding logarithm is the standard log, $\log_{10}{x}$, or simply $\log{x}$. The logarithm is the inverse of the exponential function; for example, if $x = 2$, then $10^2 = 100$. The inverse therefore (the $\log{x}$) will get us the exponent, or 2. Therefore, $\log_{10}(100) = 2.$

The base of the logarithm in this case is the $10$ from $10^x$.

More generally, if you had an exponential function $b^x$ then the corresponding logarithm would be $\log_b{x}$ in which the base is $b$.

Does this answer your question?


$\log_{16} 2^{75} = \frac {75}{4} = 18.75$ so there are $18 + 1 = 19$ digits in hex.

And indeed $2^{75} = 8*2^{72} = 8*16^{18}$ so $2^{75}$ in hexadecimal is $8000000000000000000$.

You seem to understand the base of the logorithms perfectly so far as I can tell.



As for the number of digits of a number in the decimal system. consider the following:

Let $x$ be any real number - for our convenience, assume it is positive. Then, obviously, there exists exactly a natural number $n\in\mathbb{N}$ such that: $$10^n\leq x<10^{n+1}$$

Now, if a number is between $10^n$ and $10^{n+1}$, as above, it means that it has exactly than $n+1$ digits. Imagine $x=9468$, for instance. In this case, $n=3$, since: $$10^3=1000\leq9468<10,000=10^4$$ So, we would like to express $n$ in terms of $x$, so as to have a formula that, given $x$, returns $n+1$. Since $n$ is appearing as an exponent of $10$, we think that taking $\log_{10}$ on every side is a way to "get $n$ out of the exponent". So, we have that: $$\begin{align*} &10^n\leq x<10^{n+1}\\ \Leftrightarrow&\log_{10}10^n\leq\log_{10}x<\log_{10}10^{n+1}\\ \Leftrightarrow&n\leq\log_{10}x<n+1 \end{align*}$$ So, $n=\lfloor\log_{10}x\rfloor$ which means that the number of digits of $x$ is exactly: $$n+1=\lfloor\log_{1}x\rfloor+1$$


Now, as for the logarithms, let us remind the definition:

Given $x,b>0$, $\log_bx$ is the - only - number that has the following property: $$b^{\log_bx}=x$$ So, the logarithm of $x$ with base $b$ is the only number which can be the exponent of $b$ such that $b$ to that exponent is equal to $x$.

So, given the symbol $\log_bx$, $b$ being the logarithms base means that $b$ is the number to which we may set $\log_bx$ as an exponent to get $x$. You can imagine the following game that helped me understand logarithms:

The Log-Game

Imagine two friends, Alice and Bob - hehe, these are probably the only friends we, mathematicians, have ever heard of - have a secret code to exchange messages. So, every message they want to send is being turned into a - probably very large - number. So, in our case, Alice and Bob are exchanging numbers in order to communicate.

But, in order to make it more difficult for other people to read their messages and find out the code, they use the following trick:

If $x$ is the message Alice wants to send to Bob, then, instead of this Alice sends Bob another number, which is $\log_b x$, for some base $b$ that they have pre-decided.

So, Bob receives a number $y$ that is not the message he wants, and, probably, has no meaning at all. How will he find the message? As mentioned above, Alice and Bob have pre-decided a base for the logarithms they are exchanging, so, what Bob has to do is to calculate: $$b^y=b^{\log_bx}=x$$ since, by the definition of logarithm, the only way to find $x$ given $y=\log_bx$ is to set it as an exponent to $b$. Every other, not knowing the exact value of $b$ cannot find that message.

Over-decimal Systems

We have heard about decimal system, binary system etc. But, the most usual case is that we use the usual symbols $0,1,2,\dots,9$ for such systems. What about hexadecimal system? There, we need $16$ different - distinct - symbols for our elementary digits. So, we introduce $6$ new symbols: $A,B,C,D,E,F$. These are just symbols, as the previous ones; $0,1,2,\dots,9$. There is no matter about that.

To make it a little more clear, we have decided that: $$0+1=1,\ 1+1=2,\ 2+1=3,\dots,8+1=9$$ Moreover, we have decided that $$9+1=10,\ 99+1=100$$ Or, to be more precise, the nature of our enumerating system - that the position of a digit is important to its evaluation - is a property that makes it feasible to use only some - finite in number - symbols and not infinitely many or strange combinations - see, for instance, the roman digits.

We have been used to using these symbols to represent numbers, which has driven us to the wrong conclusion that these symbols are the numbers they represent. Well, no symbol is any number, it just represents one, probably different with respect to our system.

For instance, in binary system: $$101$$ represents what in the decimal system we would write as: $$5$$ or what in the ternary (3) system we would write as: $$12$$ or what in the hexadecimal system we would write as: $$5$$

So, if we want to use some enumeration system with more that $10$ elementary digits, we have to introduce some new symbols for these digits and we also have to define the basic operations with them. So, in hexadecimal system we deicide that: $$9+1=A,\ A+1=B,\ B+1=D,\dots,E+1=F$$ and all the other properties we know.

So, to find how many elements a number $x$ has in hexadecimal, we simply note that there exists exactly on $n$ such that: $$16^n\leq x<16^{n+1}$$ and, with the same thoughts as above, we take $\log_16$ to every side, so we have: $$n\leq\log_{16}x<n+1$$ and, we have, finally that number $x$ in the hexadecimal system has: $$\lfloor\log_{16}x\rfloor+1$$ digits.


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