# Converting a cryptographic hash to a string of English words: how many words are needed? (need help with exponentials)

A particular cryptographic hash is represented as a $57$ byte string, encoded as base $64$.

RWSvUZXnw9gUb70PdeSNnpSmodCyIPJEGN1wWr+6Time1eP7KiWJ5eAM


I want to convert that string to a human-readable string, composed of dictionary words, of a similar size.

For example this string could be rendered as something like

football arrow never water


Assuming there are $10,000$ English language words, to figure out how many words it would take to represent this string, I need to solve for $x$:

$$\large (57\text{ bytes}) ^ {\text{base }64} = (x\text{ bytes}) ^ {10,000 \text{ dictionary words}}$$

My first thought on how to do this is to raise both sides to $1/64$ to get rid of the $64$ exponential

$$\large (57^{64}) ^ {1/64} = (x^{10,000}) ^ {1/64}$$

However I'm not sure to do this. Wolfram Alpha returns $x^{625/4}$, which doesn't make much sense.

How can I solve this?

The base of the representation doesn't affect how much information the string has. In other words, it takes the same amount of information to write 57 bytes in binary as it does to write those very same 57 bytes in base 64. So the right computation to do is \begin{align*} 57\text{ bytes}=456\text{ bits} \quad &\leadsto\quad 2^{456}\text{ possibilities}\\ x\text{ English words}\quad &\leadsto\quad 10000^x\text{ possibilities} \end{align*} and you want \begin{align*} 2^{456}&\leq 10000^x\\\\ 2^{456}&\leq 10^{4x}\\\\ 456&\leq 4x\log_2(10)\\\\ \frac{114}{\log_2(10)}&\leq x \end{align*} which happens for $x\geq 35$. That is, you need a string of $35$ English words to encode $57$ bytes, assuming your English words are being chosen from a fixed $10000$ word dictionary.
You are solving the wrong equation: you have $64^{57}$ not $57^{64}$ possible hashes, and the number of representable hashes using $x$ words is $10000^x$.
$$10000^x \geq 64^{57} \\ x \log(1000) \geq 57 \log(64) \\ x \geq \frac{57 \log(64)}{ \log(10000)} \approx 25.7$$ so you need key phrases of $26$ or more words.