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On a recent exam, in an integer-type question in Chemistry, I calculated my final answer to be 10^0.6, I thought I had calculated the answer incorrectly as there was no way that would be the answer. I just left it since there were no calculators allowed. Turns out, putting it into a calculator later, it's about 3.99 or 4 (That was the answer). I asked my Maths teacher but he didn't have a way (Except memorizing the anti-log table), How do you go about estimating numbers like these?

OneChannelEverything
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    Not sure what an "integer-type" question is. Do you mean to say that you know in advance that the answer is an integer? If so, then note that $10^{.6}>\sqrt {10}>3$ and $10^{3/5}<5 \iff 10^3<5^5$ and the latter is easily checked. – lulu Oct 17 '23 at 12:21
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    A short table of anti-log or log may indeed help... In this case $\log_{10}(2)\approx 0.3$ (or better $0.30103$) so that it is nearly $2*2$ – Raymond Manzoni Oct 17 '23 at 12:32
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    @RaymondManzoni Manzoni Ahh I see... I did know the log of 0.3 but I didn't think of it that way. That's quite useful and the best one I've found! – OneChannelEverything Oct 17 '23 at 13:17
  • @lulu Yes, an integer-type question does mean that the answer would be an integer. Thanks for the explanation – OneChannelEverything Oct 17 '23 at 13:20

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There is no general recipe for doing it in your head. But in this case you want $$ 10^{0.6}=10^{3/5}=(10^3)^{1/5}=\sqrt[5]{1000}. $$ Since $4^5=1024$, the number $10^{0.6}$ is very slightly below $4$.

Martin Argerami
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