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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?

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Good answers have been given below. I want to add that you should only use the differential approximation, when the variable changes by a small amount. The catch is that what "small" means depends on the function. The standard result is that $$f(x)=f(x_0)+f'(c)(x-x_0)$$ for some $c$ between $x$ and $x_0$. The question then is: how much can $f'(c)$ change within this interval? Here you see that $D(266^x)$ alone changes by a factor of 16 within this range. That speaks volumes against using this approach. Work out several examples to get a feeling of when it is ok to do this, and when not to! –  Jyrki Lahtonen Jun 11 '11 at 6:39

3 Answers 3

up vote 13 down vote accepted

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 $x_0=400$: $$f(399)\approx 5320+266\cdot\frac{1}{2\sqrt{400}}\cdot(-1)=5320-6.65 = 5313.35.$$

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4  
The advantage of joriki's approach in this particular case is that it is an underestimate, so given the choices you don't have to think about how good of an approximation it is. I gave a very slight overestimate, but no proof that it is "slight" enough to still make E. the correct choice. Along the same lines, but answering the multiple choice question with more direct certainty, you could use $\sqrt{399}\cdot 266\gt 19\cdot 266=5054$. –  Jonas Meyer Jun 4 '11 at 7:22
2  
As for the title question: $\log(266)\approx\log(100e)=2\log(10)+1$. To approximate $\log(10)$ I'll use $e\approx 2.7$, so $e^2\approx 2.7^2\approx 7.3$, and an approximation of the differential approximation is $\log(10)\approx 2+\frac{1}{7.3}\cdot 2.7\approx 2.37$ (with a quick approximate long division). So $\log(266)\approx 2\cdot2.37+1=5.74$. –  Jonas Meyer Jun 4 '11 at 7:46
2  
However, even with a more accurate version of $\log(266)$, the differential approximation indicated in the question comes out closest to A. It could be expected that the first order approximation would be bad that far away from $x_0=1$, because the second derivative of $\sqrt{x}266^x$ is large in the relevant interval. –  Jonas Meyer Jun 4 '11 at 8:00

$$\sqrt{1.5}(266)^{1.5}>\sqrt{1.44}(256)^{1.5}=1.2\cdot256\cdot16=1.2\cdot4096=4920-4.8\;.$$

Since this is closer to E than to D, the answer must be E.

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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 ${2 \over 0.5} < log_e(266) < {3 \over 0.4}$.

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@Jonas: ah right, thanks; I noticed the error yesterday but gets distracted by something else and forgot to fix it. –  Lie Ryan Jun 5 '11 at 6:02

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