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My electronics lecturer was able to instantly solve $10^{-3/20}$ as $1/2^{1/2}$, but he was not able to explain it to me because he said that it was just a number he was very familiar with.

FYI, the value $10^{-3/20}$ was derived from the gain conversion formula for ratio to decibels (dB): 20 lg (Gain ratio i.e. Vout/Vin) = Gain (dB), where gain = -3 dB.

If I put $10^{-3/20}$ into the calculator, the number looks very similar to what we get for $1/2^{1/2}$. However, I can't just assume this from the calculator.

I'd really like to know if there is a way to mathematically obtain $1/2^{1/2}$ from $10^{-3/20}$.

gradstudent
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Harmony
  • 21

3 Answers3

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$$10 ^{3\over 20}=1000 ^{1\over 20}=(1000^{1\over 10})^{1\over 2} $$ But $1000^{1\over 10}$ is approximately $2$ since $2^{10}=1024$. That is, $$10 ^{3\over 20}\approx 2^{1\over 2}\implies 10 ^{-{3\over 20}}\approx {1\over 2^{1\over 2}}$$

Fermat
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A commonly used approximation is $\log_{10}(2)\approx0.3$.

This is not exact; more precisely $\log_{10}(2)\approx0.3010299957$. However, using this approximation gives $2\approx10^{3/10}$. Raising to the $-1/2$ power yields $$ \frac1{2^{1/2}}\approx10^{-3/20} $$

robjohn
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Raising both numbers to the $20^{th}$ power,

$$10^3=1000\approx2^{10}=1024.$$

People familiar with logarithms and decibels know that $10\log_{10}2=3.0103\cdots$, i.e. $3$ for back-of-the-envelope computations.