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.

Solution 1:

$2013$ 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 $8$.

Solution 2:

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

Solution 3:

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 $x \approx 8$