How to estimate magnitude of expontent? When given an exponent, such as 6^12, is there a simple way to approximate how large(magnitude) the result is, without performing the calculation? Is this method accurate for large exponents? 
 A: One way is to learn the common logarithms of the integers 2 through 9 to one or two decimal places. So  $ \ \log_{10} 6 \ ≈ \ 0.8 \  \ \ \Rightarrow \ \  \log_{10}  \ 6^{12} \ ≈ \ 9.6 \ $ . Since  $ \ \log_{10} 4 \ ≈ \ 0.6 \ $   , we can estimate $ \  6^{12} \ ∼ \ 4 ⋅ 10^9 \ $ (which turns out to be somewhat high -- by about a factor of 2). 
Another way might be to use $ \ 6^3 \ = \ 216 \ ≈ \ 200 \ $ , so  $ \ 6^{12} \ ≈ \ 200^4 \ = \ 2^4 ⋅ 10^8 \ = \ 1.6 ⋅ 10^9 \ $ (a little low).  If you work with numbers much, you will find it helpful to learn (actually, you start to involuntarily memorize) the low integer powers of the low integers.  (Over the years, I've ended up with "times-tables" out past 20 x 20 , and powers of the integers of at least 2 through 20 up to cubes or more...)
A: When you are doing estimation, you need to be conscious of the allowable errors.  Knowing some logs helps, too. If you know that $\log_{10}2 \approx 0.3$ and $3^2\approx 10$ you are done. $\log_{10}6^{12}=\log_{10}(2^{12})+\log_{10}(3^{12})\approx 12 \cdot 0.3+6=9.6,$ so $6^{12}\approx 4\cdot 10^9$  Douglas Hofstadter claimed that this is the region we care mostly about how many zeros there are.  Of course, you can get the exact result $2176782336$, which is a factor $2$ below my approximation-mostly because $3^2=9$, not $10$, but I claim a factor $2$ here is acceptable.  Your mileage may vary.
