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I encounter such a problem, in a Maths contest, to find out the closest integer to $\ln(2013)$, without using a calculator. I really get stuck.

I tried to turn $\ln(2013)$ into $\ln(3)+\ln(11)+\ln(61)$, but nothing valuable obtained. I applied also Taylor series of natural log but it doesn't work. Any suggestions are welcomed.

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I thought this was a rather weird question as given, and seeing the different answers posted and how all of them use the damn symbol $\,\approx\,$ , I think it should/must be given some rational approximations to some values of the logarithm, like $\,\log 2\approx 0.7\,$ and etc. Without this assumption anyone can use almost any "approximation" and things get murky... –  DonAntonio Feb 28 '13 at 11:58
    
Thanks for your opinion. I know $\ln(2)$. But it's really unexpected that they use $\ln(10)$ and $e^3$. I ask the question in this forum now and naturally you find it weird. However it's a question in a contest, so I realise every single method will do, as long as they can remember the estimation. For me the simpler the estimation used is, the nicer the question is. –  Michael Li Feb 28 '13 at 14:14
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@DonAntonio I think the ln(2)≈0.7 approximation is adequate in this case, because we are looking for an integer, we need only one (or at most 2) significant digits for the computation. –  RudolphEst Feb 28 '13 at 15:10
    
This can clearly be solved simply by knowing the approximate value of $e$ and multiplying decimals (see Alex Jordan's answer). It is remarkable to me that this simple approach is not considered obvious; perhaps multiplying decimals is no longer something that is ever done "without using a calculator"! –  David Bevan Mar 7 '13 at 9:21

10 Answers 10

up vote 22 down vote accepted

$2013$ is "very" close to $2048=2^{11}$. So how about $$2013=e^x=2^y$$ where $y$ is effectively equal to $11$. Then $x=y\ln 2$ and $\ln 2$ is famously equal to $0.7$. Then $$\ln(2013)\approx 11\cdot 0.7=7.7$$ giving an answer of $8$.

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Oh drats. I wrote a similar argument. Only that $\ln 2$ is not famously equal to any rational number, in particular not to $0.7$. –  Asaf Karagila Feb 28 '13 at 6:08
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right, it's a joke... –  Jonathan Feb 28 '13 at 6:09
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Sort of like how $\pi^4 + \pi^5 = e^6$, etc, I guess: en.wikipedia.org/wiki/… –  Neal Feb 28 '13 at 6:15
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I've memorized $\ln 2$ to 3 decimal places. –  Joe Z. Feb 28 '13 at 12:59
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This method seems a bit dangerous, given the 'correct' answer turns out to be less than 7.61. We wouldn't want to get too close to 7.5 –  Thomas Ahle May 8 '13 at 12:35

Without remembering logs (while it may be useful to recall some),
Note $2 < e < 3$ and $2^{10} < 2013 < 3^7$
So if $e^x = 2013$, we must have $7 < x < 10$

Hence $e^\frac{x}{11} = 2013^\frac{1}{11} = (2048 - 35)^\frac{1}{11} = 2(1 - \frac{35}{2048})^\frac{1}{11}$

Now we have $\frac{x}{11} < 1$ and can approximate without fear of losing much accuracy using:

$1 + \dfrac{x}{11} + \dfrac{x^2}{242} \approx 2 - \dfrac{2\cdot 35}{11 \cdot 2048} $

leading to
$x^2 + 22 x \approx 241$
$(x+11)^2 \approx 362$
or $x \approx 8$

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Neat method, but quite complicated comparing to remembering that $\ln 2\approx 0.691$. –  Asaf Karagila Feb 28 '13 at 7:18
    
Agreed. Had a friend who memorised log and antilogs of many numbers, which was quite useful, till calculators became common. Still remembering a few is good, and so is Pseudonym's trick. –  Macavity Feb 28 '13 at 7:28

Note that $2013$ is nearly $2048$ which is $2^{11}$.

Also note that $\ln(2013)=\log_2(2013)\cdot\ln 2$. Since $\log_2(2013)$ is nearly $\log_2(2048)=11$ and $\ln 2$ is roughly $0.693\approx 0.7$ we have that $\ln(2013)$ is roughly $11\cdot0.7\approx 7.7\approx 8$.

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Another useful rule of thumb is that to within 1%, $\ln x + \log_{10} x \approxeq \log_2 x$. So $\log_2 2013 \approxeq 11$ and $\log_{10} 2013 \approxeq \log_{10} 2000 = 3 + \log_{10} 2 \approxeq 3.3$. So $\ln 2013 \approxeq 7.7$. If you've forgotten an approximation for $\log_{10} 2$, you can use the fact that $\sqrt{\sqrt{10}} \approxeq 1.78$, so $\log_{10} 1.78 \approxeq 0.25$. –  Pseudonym Feb 28 '13 at 6:17

$\ln 3$ is a little greater than $1$. In fact you can use $\ln (1+x)\approx x$ with $\frac 3e \approx 1.1$ to get $\ln 3 \approx 1+ \ln 1.1 \approx 1.1$

Maybe you know that $\ln 10 \approx 2.3$, so $\ln 11=\ln 10 + \ln 1.1 \approx 2.4$

Then $\ln 61 \approx \ln 2 + \ln 3 + \ln 10 \approx 0.7+1.1+2.3 =4.1$.

Summing it all up, we have $1.1+2.4+0.7+1.1+2.3=7.6$ and I would say $8$, though we were uncomfortably close to $7.5$. In fact $\ln 2013 \approx 7.607$ so the approximations were quite good.

Afterthought: even easier is $\ln 2000 \approx 0.7+3\cdot 2.3 = 7.6$ and the extra factor $1.006$ only adds $0.006$

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One of my useful memorized rough approximations is $e^3\approx20$, and I know that $20$ is a slight underestimate. So $e^6$ is a bit over $400$, and $e^9$ is a bit over $8000$. That means that the choice is between $7$ and $8$. $400$ is too small by a factor of about $5$, and $8000$ is too big by a factor of only about $4$, so it’s $8$, though not by a whole lot. (And sure enough, it turns out to be about $7.61$.)

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Clearly the answer is not all that large. So why not just multiply out powers of $e$ to a reasonable number of digits, like 3? Without too much work you'll hit close to $2013$ soon enough. All calculations below are by hand with either two or three digits preserved.

$$e\approx2.72$$ $$e^2\approx7.40$$ $$e^4=(e^2)^2\approx54.8$$ $$e^8=(e^4)^2\approx3000$$

Back pedal

$$e^6=(e^4)(e^2)\approx406$$ $$e^7=(e^6)(e)\approx1100$$

So with $e^7\approx1100$ and $e^8\approx3000$ we must judge where $2013$ falls.

$$1100\rightarrow(\times\approx1.8)\rightarrow2013\rightarrow(\times\approx1.5)\rightarrow3000$$ shows us that 2013 is relatively closer to 3000 than 1100. So we'd say the answer is 8. A formal proof would require more care paid to error bounds on all of the estimation ($\approx$).

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Another approach to the second half is to memorize the value of $e^{1/2}$ and use that to locate the right exponent. –  Thomas Feb 28 '13 at 9:18
    
That would work. It feels like that would be an even less likely thing to have memorized than some of the log values in other answers. I was shooting for an approach that used more common trivia. –  alex.jordan Feb 28 '13 at 19:27
    
Quite true, though you can always approximate it with a couple Newton-Raphson iterations, so you don't strictly have to memorize it. –  Thomas Mar 1 '13 at 0:53

Well $\,\ln(2)\approx 0.69315\,$ and $\,\ln(10)\approx 2.302585\ $ so that : $$\ln(2013)=\ln(2)+\ln(10^3)+\ln(1.0065)\approx 0.69315+3\cdot 2.302585+0.0065$$ (with an error of order $\frac 12 0.0065^2$)
getting : $$\ln(2013)\approx 7.6074$$ (of course $\ \ln(2013)\approx 0.7+3\cdot 2.3\approx 7.6\ $ was enough here...)

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$2013$ is "very" close to $2000=2\cdot10^3$. So how about $$\ln(2013)\approx \ln(2000)$$ $$\ln(2000) = \frac{\log(2000)}{\log(e)}$$ remembering $$\log(e) = \frac{1}{\ln(10)}$$ then we have $$\ln(2000) = \log(2000)\cdot\ln(10)$$ $$ = \log(2\cdot 10^3)\cdot\ln(10)$$ $$ = (\log(2) + 3\cdot\log(10))\cdot\ln(10)$$ $$ \approx (0.3 + 3)\cdot 2.3 $$ $$ = 3.3 \cdot 2.3 $$ $$ \approx 7.6 $$ $$ \approx 8 $$

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Look at integer powers of $3$. We have $3^6=729$, so $3^7$ is about $2200$, bigger than $2013$.

For $e$, which is about $2.7$, we may need a bigger exponent, maybe $8$ or even $9$. We have that $3^8$ is about $6500$. And $(0.9)^8$ is therefore roughly $4\times 10^{-1}$. Multiply by $6500$. This puts us over $2013$. So exponent $8$ is too big, but closer than $7$.

Remark: In hindsight I should have worked directly with $2.7$. But the post describes how I actually calculated.

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If you remember that $e^3 \approx 20$:

$$ln(2013) \approx ln(2000) = ln(20 \cdot 20 \cdot 5) = ln(20) + ln(20) + ln(5) \approx 3 + 3 + ln(5) $$

ln(5) is between 1 and 2 (because $e \approx 2.71$ and $e^2 \approx 7.4$), so all you need to known is if $ln(5)$ is greater or less than 1.5.

$e^{1.5} = \sqrt{e^3} \approx \sqrt{20} = \sqrt{4 \cdot 5} = 2 \cdot \sqrt{5} \approx 2 \cdot 2.2 = 4.4 < 5$, so $ln(5) > 1.5$.

=> $ln(2013) \approx 8$.

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