Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@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 $8$.

share|cite|improve this answer
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
right, it's a joke... – Jonathan Feb 28 '13 at 6:09
Sort of like how $\pi^4 + \pi^5 = e^6$, etc, I guess:… – Neal Feb 28 '13 at 6:15
I've memorized $\ln 2$ to 3 decimal places. – Joe Z. Feb 28 '13 at 12:59
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

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 $11\cdot0.7\approx 7.7\approx 8$.

share|cite|improve this answer
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

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 $x \approx 8$

share|cite|improve this answer
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

$\ln 3$ 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 $0.006$

share|cite|improve this answer

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 $7.61$.)

share|cite|improve this answer

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 ($\approx$).

share|cite|improve this answer
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

$2013$ 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 $$\ln(2000) = \log(2000)\cdot\ln(10)$$ $$ = \log(2\cdot 10^3)\cdot\ln(10)$$ $$ = (\log(2) + 3\cdot\log(10))\cdot\ln(10)$$ $$ \approx (0.3 + 3)\cdot 2.3 $$ $$ = 3.3 \cdot 2.3 $$ $$ \approx 7.6 $$ $$ \approx 8 $$

share|cite|improve this answer

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...)

share|cite|improve this answer

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.

share|cite|improve this answer

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 $ln(5) > 1.5$.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.