Quick preface as with my other posts: I'm not a math major, but I'm enjoying the math portion of my studies, so please keep any explanation on an idiot's level lol - thanks!
I came across a problem yesterday that had an exponential value for the answer of something like $3^{79}$. This is obviously no problem for a calculator even with the solution set in base 10. I was trying to figure out how to algebraically manipulate this to solve or approximately estimate a solution with a base of 10 without using a calculator.
I saw a post that suggested an estimation between base 2 and base 10 by using $2^{10} \approx 10^3 : 1024 \approx 1000$.
Consider $$3^{79}=\left(10^{\log_{10}3}\right)^{79}=10^{79\log_{10}3}.$$ The exact answer would thus be to just multiply your exponent by $\log_{10}3$. For quick mental approximations, it is good to have identities like $2^{10}\approx 10^3$, since that is easier to remember than the value of $\log_{10}2$. For $3$, you can start with $3^{2k}\approx 10^k$, even though the error is quite large. A better approximation is $3^{21}=10^{10}$, but the numbers are getting quite large at that point.