I had the following question:
Given a number in $base^{exponent}$ format, like $3^{500}$ or $7^{300}$, is
there a way to convert it to scientific notation without explicitly
constructing the (possibly very large) number as an intermediate step? We will
assume here that both $base$ and $exponent$ are positive. For example,
$7^{300}$ should be converted to $3.384\ldots \cdot 10^{253}$, also written
3.384...e+253.
This can be useful when looking at exponential growth with different bases. I happened to come across this particular question, right now, with game tree branching factors. If one game has an average branching factor of 3 and lasts 500 turns on average, is that more or fewer total states than one with an average branching factor of 7, which lasts 300 turns on average? We can eyeball the answer easily by converting to scientific notation, with its more familiar base 10: $3^{500} \approx 3.636 \cdot 10^{238}$, while $7^{300} \approx 3.384 \cdot 10^{253}$, so the latter is larger by 15 orders of magnitude.
Beyond games, exponential growth is recently in the news in epidemiology of course. Given an effective reproduction number $R$ and current cases $i$, there will be an expected $i \cdot R^t$ cases after $t$ additional time units pass. But how to easily compare different $R$ and $t$ combinations? Converting to scientific notation is one way.
I spent more time googling unsuccessfully for this answer than it ultimately took to just derive it from equations solvable using high school mathematics. Maybe I am bad at Google. But nonetheless, I thought I'd summarize the result.
There are two steps. First, change bases from $base$ to $10$. The resulting exponent is likely not to be an integer. Secondly, take this $10^x$ and convert it to scientific notation, meaning an expression of the form $m \cdot 10^e$, for some mantissa $m$, $0 < m < 10$ and integral exponent $e$.
We want to turn $base^{exponent}$ into $10^x$. That means solving for $x$ in the equation $10^x = base^{exponent}$. For example, $10^x = 7^{300}$.
First, take the base-10 logarithm of both sides: $$\log_{10} 10^x = log_{10} base^{exponent}$$ Applying the definition of a logarithm as the inverse of exponentiation, this is equivalent to: $$x = \log_{10} base^{exponent}$$ This gives us a closed form for $x$, but still requires evaluating $base^{exponent}$, which is what we were hoping to avoid. But we can apply the logarithmic identity that $\log a^b = b \cdot \log a$, resulting in: $$x = exponent \cdot \log_{10} base$$
Now we have something that can be computed efficiently, at least in approximate terms. For example, we can convert $7^{300}$ to $10^x$ by computing $x = 300 \cdot \log_{10} 7$, yielding $10^{253.529\ldots}$.
This is almost scientific notation, except that scientific notation has integral exponents, not exponents like $253.529\ldots$.
We now have a $10^x$ with probably non-integral $x$, such as the running example of $7^{300} \approx 10^{253.529}$.
The exponent for the purposes of scientific notation will simply be this $x$ rounded down to the nearest integer, denoted $\lfloor x \rfloor$. That will naturally result in a smaller number. So to approximate the original number, we need to multiply it by a mantissa $m$. This is the basic idea of scientific notation: a number between 0 and 10 times an integral power of 10. Put in the form of an equation, we want to solve for $m$ in: $$m \cdot 10^{\lfloor x \rfloor} = 10^x$$ This can be solved with high school mathematics actually a little more elementary than Step 1, though for some reason it took me longer to recall that fact: $$m = \frac{10^x}{10^{\lfloor x \rfloor}}$$ $$m = 10^{x - \lfloor x \rfloor}$$ Now we again have a closed-form equation that can be computed reasonably efficiently with floating-point arithmetic (not requiring a gigantic intermediate result).
To return again to our running example of $7^{300}$. In step 1 we determined this was equal to $10^{253.529\ldots}$. Now we turn it into scientific notation of the form $m \cdot 10^e$, where:
3.384e253,
without explicitly constructing the large intermediate number.
import math base = 7 exponent = 300 raw_exponent = exponent * math.log10(base) sci_notation_exponent = math.floor(raw_exponent) sci_notation_mantissa = 10 ** (raw_exponent - sci_notation_exponent)