Find the integer closest to $\ln(2013)$ 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.
 A: $\ln 3$ 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 $0.006$
A: 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 $7.61$.)
A: 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 ($\approx$).
A: 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...)
A: $2013$ 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
$$\ln(2000) =  \log(2000)\cdot\ln(10)$$
$$ = \log(2\cdot 10^3)\cdot\ln(10)$$
$$ = (\log(2) + 3\cdot\log(10))\cdot\ln(10)$$
$$ \approx (0.3 + 3)\cdot 2.3 $$
$$ = 3.3 \cdot 2.3 $$
$$ \approx 7.6 $$
$$ \approx 8 $$
A: $2013$ 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 $8$.
A: 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. 
A: 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 $11\cdot0.7\approx 7.7\approx 8$.
A: 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 $x \approx 8$
A: 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 $ln(5) > 1.5$.
=> $ln(2013) \approx 8$.
