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?