the approximation of $\log(266)$? Consider the following exercise:

Of the following, which is the best
  approximation of
  $\sqrt{1.5}(266)^{1.5}$?
A 1,000 B 2,700 C 3,200 D 4,100 E
  5,300

The direct idea is using the "differential approximation":
$$f(x)\approx f(x_0)+f'(x_0)(x-x_0)$$
where $f(x)=\sqrt{x}266^x$ and $x_0=1$, $x=1.5$. 
Finally, one may have to approximate $\log 266$. So here are my questions:


*

*How to approximate $\log 266$?

*Is there any other methods to answer this question?
 A: $$\sqrt{1.5}(266)^{1.5}>\sqrt{1.44}(256)^{1.5}=1.2\cdot256\cdot16=1.2\cdot4096=4920-4.8\;.$$
Since this is closer to E than to D, the answer must be E.
A: I would use $\sqrt{1.5}\cdot 266^{1.5}=\sqrt{1.5}\cdot\sqrt{266}\cdot 266=\sqrt{399}\cdot 266\approx\sqrt{400}\cdot 266=20\cdot 266 = 5320$.
You could make this more accurate using the differential approximation of $f(x)=266\sqrt{x}$ at $x_0=400$: $$f(399)\approx 5320+266\cdot\frac{1}{2\sqrt{400}}\cdot(-1)=5320-6.65 = 5313.35.$$
A: assuming that log is a log in base e, then $log_e 266 = {{log_{10} 266} \over {log_{10} e}}$. Now, approximating $log_{10}$ is easy, just count the number of digits in 266 and subtract 1, so $log_{10} 266 = log (2.66 \cdot 10^2) \approx 2$. Now, you have to know that $log_{10} e = 0.43429448190325176 \approx 0.5$. So ${{log_{10} 266} \over {log_{10} e}} \approx {2 \over 0.5} \approx 4$.
If you want a more rigorous approximation, then we start with the same argument $log_e 266 = {log_{10} 266 \over log_{10} e}$; then let $x = log_{10} 266$ therefore $2 < x < 3$; so since $y = log_{10} e$ is between $0.4 < y < 0.5$, therefore since $log(266) = x/y$ then ${2 \over 0.5} < log_e(266) < {3 \over 0.4}$.
