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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?

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    $\begingroup$ One way is to learn the common logarithms of the integers 2 through 9 to one or two decimal places. So $ \ \log_{10} 6 \ \approx \ 0.8 \ \ \Rightarrow \ \ \log_{10} 6^{12} \ \approx \ 9.6 \ . $ Since $ \ \log_{10} 4 \ \approx \ 0.6 \ \ $ , we can estimate $ \ 6^{12} \ \sim \ 4 \ \cdot \ 10^9 \ $ (which turns out to be somewhat high -- by about a factor of 2). Another way might be to use $ \ 6^3 \ = \ 216 \ \approx \ 200 \ $ , so $ \ 6^{12} \ \approx \ 200^4 \ = \ 2^4 \ \cdot \ 10^8 \ = \ 1.6 \ \cdot \ 10^9 \ $ (a little low). $\endgroup$ Commented Jul 29, 2015 at 3:27
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    $\begingroup$ @RecklessReckoner: that should be an answer. $\endgroup$ Commented Jul 29, 2015 at 3:32
  • $\begingroup$ Estimation is made much easier as you have more facts available without thinking about them. $\log_{10}2 \approx 0.3 \implies 2^{10} \approx 1000$ is amazingly useful. That plus $(1+x)^n \approx 1+nx$ for $x\ll1$ can get you a long ways. More facts and more Taylor series make it easier. Like so many other skills, if you want to do this well, you need to practice. $\endgroup$ Commented Jul 29, 2015 at 4:02
  • $\begingroup$ As an example, what is the volume of the earth? The radius is 4000 miles or 6400 km, which is $2^6\cdot 10^2$ km. The volume is $\frac 43\pi r^3$ and (since $\pi=3$ for this purpose) $V=4(2^6 \cdot 10^2)^3=2^{20}10^6=10^{12}$km$^3$ $\endgroup$ Commented Jul 29, 2015 at 4:07

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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...)

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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.

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    $\begingroup$ Estimation is an under-taught, under-utilized skill in the age of cheap calculators. But it is quite often valuable for making basic judgements where an order of magnitude (or half-order) is adequate for a decision, or for doing simple "sanity-checks" of all manner of calculations. I tell students they should learn ways to estimate answers first, and use calculators or computers only when "precision" work is needed. $\endgroup$ Commented Jul 29, 2015 at 3:52

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