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What is the method for mentally computing an approximation of $3^{3.5}$ and similar calculations? (without using any calculator)

The best I did is: $3^{3.6}=e^{3.6ln(3)}=e^{3.6*1.098} \approx e^{3.6*1.1}=e^{3.96} \approx e^4=54.5$

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  • $\begingroup$ This is your third very recent "how do I find ... without a calculator?" question. Consider searching for "mental arithmetic" to find books and websites that offer the kinds of strategies you seek. $\endgroup$ – Ethan Bolker Dec 20 '17 at 15:48
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    $\begingroup$ Well, you know it will be above $3^3=27$ and you know it will be less than $3^4=81$. You should also know that $1<3^{0.5}<4^{0.5}=2$... with this information alone you could say $3^{3.5}$ is a bit less than twice $3^3$ and so is a bit less than $54$. You can get more accurate by knowing the first few digits of the decimal representation of $\sqrt{3}\approx 1.73$, putting $3^{3.5}$ as almost three quarters larger than $27$, so closer to $27+21=48$. (true value is approximately $46.765$) $\endgroup$ – JMoravitz Dec 20 '17 at 15:48
  • $\begingroup$ You seem to equate mental calculation with without a calculator. Paper and pencil is a significant step in between. I had prepared an answer for your $x^x=100$ that was hopeless mentally but was achievable (with work) on pencil and paper. Which do you want? $\endgroup$ – Ross Millikan Dec 20 '17 at 15:50
  • $\begingroup$ How did you mentally compute that $\ln(3)\approx 1.098$? And why do you show three digits of precision in your answer when it is easy to show (because $e^{-0.04}\approx 1-0.04$) that only the first digit is correct? $\endgroup$ – David K Dec 20 '17 at 15:55
  • $\begingroup$ I knew ln(3) as I know some of the first ln(p) where p is prime. I know as well exp(x) for x integer from 1 to 10. I did all the computations mentally. $\endgroup$ – astudentofmaths Dec 20 '17 at 17:05
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Methods for mental calculation often depend on knowing facts. For this problem, knowing $\sqrt 3\approx 1.732$ makes it easy. $$3^{3.5}\approx 3^3\sqrt 3=27\cdot 1.732=46.764$$ where I did the final multiply mentally and I should delete the last digit because I only have four digits accuracy in the input. It is also important to be extremely facile with $(1+x)^n \approx 1+nx$ for $nx \ll 1$. This lets you adapt things to the facts you know.

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Memorize three logarithms

Here’s how I do it. I’ve got the base-$10$ logarithms for the integers $2$, $3$, and $7$ memorized:

$$\begin{array}{|r|r|} \hline n & \log n \\ \hline 2 & .301 \\ \hline 3 & .477 \\ \hline 7 & .845 \\ \hline \end{array}$$

That’s all you need to get the logarithm for the first digit of any number, because you can easily deduce the logs for the other first digits from $.301$ and $.477$. For example, $\log 6 = \log 2 + \log 3 \approx 0.301 + 0.477 = 0.778$.

So, I can use logarithms to reduce exponentiation to multiplication, which I can do quickly in my head if I only consider the first couple digits. First I remember:

$$\log 3 \approx 0.477$$

$0.477$ is a little less than half, and half of $3.5$ is $1.75$, so here’s another quick mental approximation:

$$3.5 \times 0.477 \approx 1.7$$

Now I just need to figure out the antilogarithm of $1.7$. The characteristic (the part to left of the decimal) is $1$, so the antilog is in the $10$’s, i.e. in the range $[10, 100)$. The mantissa is $0.7$, which I recognize as just about $0.301$ less than $1$. Referring to the memorized table again, $0.301 \approx \log 2$, so $10^{0.7} \approx 10 \div 2 = 5$. That gives me the first digit of the answer. So, I conclude:

$$3^{3.5} \approx 50$$

A calculator reveals that the true answer is $46.765372$. So, not bad for a few seconds of mental fiddling.

The trade-off

Notice that Ross Millikan’s approximation is much more accurate. His approach relies on having a large library of exploitable facts memorized, and capitalizing on the most relevant ones. That’s also how lightning calculators do it. The method in this answer is crude but very easy to learn. You just memorize three numbers and you’re set for life. It was told to me by a chemistry professor long ago. It’s good for quickly estimating a great many calculations in physics, chemistry, and the like that involve exponents or roots. Sometimes it can be hard to estimate the antilogarithm. I suppose I could memorize that $10^{0.1} \approx 1.26$ and maybe a couple more, but I never have.

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  • $\begingroup$ I've edited your answer so that the layout is more consistent. (same font for numbers in the table as outside the table.) You may want to see this MathMeta.SE question: math.meta.stackexchange.com/q/4240/290189 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 20 '17 at 17:13
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    $\begingroup$ @GNUSupporter Thanks! For both the fix and the link. Of course, make the table with LaTeX. Why didn't I think of that? :) $\endgroup$ – Ben Kovitz Dec 20 '17 at 17:19
  • $\begingroup$ In the square roots, I only have 2,3,10 memorized to three places each. One can derive a lot from them. I was lucky this problem called for one. The base 10 logs of 2,3 are very useful as are natural logs of 2,10. $\endgroup$ – Ross Millikan Dec 20 '17 at 21:04
  • $\begingroup$ @RossMillikan Ah, now I see that the approach in your answer requires less background knowledge than I thought (and less "luck"). And to be fair, the "tiny log table in your head" method probably still requires one to memorize $\sqrt 10$ to get the antilog of $0.5$. BTW, the chemistry prof actually said that all you need to memorize is $\log 2 \approx 0.301$ and you can get all the rest from that. Any idea what he meant by that? I have indeed found that $\log 2$ is the most useful, but how would one get $\log 3$ and $\log 7$? $\endgroup$ – Ben Kovitz Dec 21 '17 at 21:16
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    $\begingroup$ I would suggest that $\log 7$ is not needed for mental calculation because not many things are divisible by that. I would just use $6$ and $8$ which you can get from $2$ and $3$. You can use $2^8=256 \approx 243=3^5$ to get $\log 3 \approx \frac 85 \log 2 \approx 0.4816$ which is pretty close. $\endgroup$ – Ross Millikan Dec 21 '17 at 21:23
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Based on the facts you had already memorized, \begin{multline} 3^{3.5} = e^{3.5\ln3} \approx e^{3.5(1.098)} = e^{3.5(1.1 - 0.002)} \\ = e^{3.85 - 0.007} = e^{3.843} = e^4 e^{-0.157} \approx 54.5 \times 0.843 = 45.9435 \approx 46, \end{multline} where the factor $0.843$ comes from the Taylor expansion of $e^x$ up to the term in $x$ (discarding $x^2$ and later terms), so it is an underestimate. Guessing that we have at most two significant terms, we round up. (You may also take some shortcuts with $54.5 \times 0.843,$ which is hard to do mentally; but try to round upward.)

If we also take the $x^2$ term of the Taylor expansion, we get $$e^{-0.157} \approx 1 - 0.157 + \frac{0.157^2}2 \approx 1 - 0.157 + \frac{0.025}2 \approx 0.855,$$ and the final approximation is $3^{3.5}\approx 54.5\times 0.855 = 46.6$ (or whatever you use to approximate $54.5\times 0.855$). That's still not quite as good (or even as easy to do) as Ross Millikan's method, but it generalizes well.

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