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Given the following expressions:

0.3² = 0.09
0.025² = 0.000625

I can head count 3*3 and 25*25, which helps me solve the above expressions. But how do I know how many 0s there should be before "9" or "625"?

It seems to me that I need to add them until I get the the number of decimal cases of the base times the exponent (1*2 and 3*2, respectively). Is that right? And what is the rationale behind that?

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closed as off-topic by José Carlos Santos, Claude Leibovici, Aqua, Shailesh, Leucippus Oct 29 '17 at 1:13

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  • $\begingroup$ In first case you have 1 zero so squaring you get 2,in the second case you have 2 zeroes and squaring you vet, if you have n zeroes then you have 2n zeroes. $\endgroup$ – kingW3 Oct 28 '17 at 8:19
  • $\begingroup$ Why does not $30^{2}=90$? The numer of zeroes represents an integral power of base $10$ and this power doubles up during squaring. $\endgroup$ – Paramanand Singh Oct 28 '17 at 8:21
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Since $0.3=3\times\frac1{10}$,$$0.3^2=\left(3\times\frac1{10}\right)^2=9\times\frac1{100}=0.09.$$And, since $0.025=25\times\frac1{1000}$,$$0.025^2=625\times\frac1{1000000}=0.000625.$$

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You can write it like this: $$ 0.3^2 = (3 \cdot 10^{-1})^2 = 3^2 \cdot 10^{-1 \cdot 2} = 9 \cdot 10^{-2} = 0.09 $$

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