Especially when working with microprocessors I frequently have to figure out how many bits I would need to capture a given value. It so happened to be that one can easily derrive an answer to that question through the expression $\lfloor \log_{b}(q) \rfloor + 1$ where $q$ is the value we want to capture and $b$ is the number of the base we want to express our value in.

This is a pretty basic expression, but I got to writing this post while compiling a list of problems to conveniently solve with bc and catching myself dumping too much details in that post.

Solving this in bc is discussed in that post with bc commands :wink: but to save you a click here goes… in case you need to figure out how many bits you will need (base 2) for the value $8$ use the following command to pipe the statements to bc for processing, needless to say that you need to change $8$ for whatever value want to represent, and $2$ for whatever the intended base of your output should be.

echo "f=l(8)/l(2); scale=0; 1+(f/1)" | bc -l
#echo "p=l(8)/l(2); scale=0; print 1+(f/1), \" bits\"" | bc -l

How many digits are needed to represent a number in base $b$?

The expression $\lfloor \log_{b}(q) \rfloor + 1$ is the answer to that problem. Working out $\log_{b}(q)$ in a calculator is simply done by calculating $\frac{\log(q)}{\log(b)}$ where $q$ is the number we want to capture and $b$ is the base of the system we want to represent that value in.

Just to rewind back to a somewhat simpler scenario, let’s think base $10$ for a second. We can capture 10 values by just using one decimal place which are $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$. That is quite simple :wink:. For every decimal place we have to assume a factor of $10^{N}$ where $N$ represents the place of decimal. The least significant digit is the one furthest to the right such that $47$ equals $(4\times 10^{1})+(7\times 10^{0}) = 40 + 7$.

$\log_{10}(100) = 2$ means that the value $100$ is captured by the second power to the base $10^2$. Ones are the zeroeth power to the base ($10^0=1$), tens are the first power ($10^1=10$) and hundredths are the second power ($10^2=100$).

We need three powers to ten ($10^3$) to represent $1000$, that means 4 digits because the power to zero covers the ones :wink:. If I ask how many decimal places are needed to represent $978$, you would answer that we need $\log_{10}(978) \approx 2.99034$ powers to the ten to cover it.

After calculating the powers needed to achieve your target value round down the result to the nearest whole number (we floor it) because we only care about the range the number is in. It is in the tens, hundreds, thousands, millions, billions, etc. Don’t care about the details, we are generalizing. For the values $100$, $500$ and $999$, which all lie within the hundreds range, the powers to the tenth base are respectively $log_{10}(100) = 2$, $log_{10}(500) \approx 2.69897$ and $log_{10}(999) \approx 2.99957$. The whole number we can round it down to is $2$ (hundreds) which gives us 3 decimal places $10^0$, $10^1$ and $10^2$.

The number of digits needed to represent the number equals the next smallest integer greater than the floor of the logarithm of the number of interest to with its base set to the base of the output value we are looking for.

$\lfloor \log_{10}(q) \rfloor + 1$

In the binary world the trick remains the same with a minor difference, the base changes. If we need 8 represented we determine that we need three ($\log_{2}(8) = 3$) powers to two to represent that number. Accounting for the zeroeth power (the ones) we get four digits and yes $1000_{2} is 8_{10}$.

$\log_{2}{8}$ $\log_{10}(47) = _{}$

The value four is represented as the third decimal digit $100^{10} = 4_{10}$