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How to find an approximation for value of $2^x$ without using calculator?

For example, $2^{4.3}$.

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    $\begingroup$ If you have the integer powers of 2 memorized, you have a rough ballpark, e.g., $16 < 2^{4.3} < 32$. Noting that three tenths is a little more than a quarter, you can improve that to $2^{4.3} \approx 20$. $\endgroup$ – Robert Soupe Apr 8 '17 at 20:27
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I suppose you are asking how to do it without a calculator or logarithm tables with a high precision. And how to do it fast. Of course, there is a way. Nowadays we calculate even indefinite integrals with computers. Well, here is what I suggest in this particular case:

$$2^{4.3}=2^3\cdot2^{1.3}=8\cdot2^{1.3}$$ Now it is very easy to get a very precise answer fast by using the Taylor's formula for fraction powers of the binomial (1+x):

$$(1+x)^{t}=1+\frac{t}{1!}\,x+\frac{t\,(t-1)}{2!}\,x^2+...+\frac{t\,(t-1)\cdot...\cdot t\,(t-n+1)}{n!}\,x^n+...$$

So, we get this:

$$2^{1.3}=(1+1)^{1.3}=1+\frac{1.3}{1!}\cdot1+\frac{1.3\cdot(1.3-1)}{2!}\cdot1^2+\frac{1.3\cdot(1.3-1)\,(1.3-2)}{3!}\cdot1^3+...$$

or

$$1+\frac{1.3}{1!}+\frac{1.3\cdot(1.3-1)}{2!}+\frac{1.3\cdot(1.3-1)\,(1.3-2)}{3!}+\frac{1.3\cdot(1.3-1)\,(1.3-2)\,(1.3-3)}{4!}...$$

For simplicity of my crude calculations I'm multiplying everything by $1000$:

$$1000+1300+\frac{130\cdot(13-10)}{2!}+\frac{13\cdot(13-10)\,(13-20)}{3!}+\frac{1.3\cdot(13-10)\,(13-20)\,(13-30)}{4!}...$$

Now we get this: $2300+195-13\cdot49/6+1.3\cdot49\cdot27/24≈2495-650/6+1.3\cdot2\cdot27≈2500-110+70=2460$

Now $\;8\cdot2460/1000=19680/1000=19.68\quad$ The actual number is 19.698... My calculation here is too crude but you get the idea. I don't even bothered with precise arithmetic here.

As the factorial in the denominator grows larger you get your precision. Note how the terms alternate (you either add a term or subtract it). I chose the number $1.3$ on a hunch to sorta "make" a few terms smallish.

I think it is important to understand the ideas, rather than memorize things like this: $$\sqrt[3]{1+x}=1+\frac{1}{3}\,x-\frac{2}{2!\cdot3^2}\,x^2+\frac{2\cdot5}{3!\cdot3^3}\, x^3-\frac{2\cdot5\cdot8}{4!\cdot3^4}\, x^4+...$$

PS: You can also calculate sines, cosines, logarithms etc. in a similar fashion that is "on paper" without logarithm tables or calculators.

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    $\begingroup$ As this is a pre-calculus question perhaps you should explicitely state what Taylors formula is. $\endgroup$ – fleablood Apr 8 '17 at 21:50
  • $\begingroup$ Yeah, you're right, thanks. I've just added it to the post. $\endgroup$ – Ken Draco Apr 8 '17 at 22:25
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Consider this: $$ 2^x \approx \dfrac{2^{\lfloor x \rfloor} + 2^{\lceil x \rceil}}{2} = 2^{\lfloor x \rfloor - 1} + 2^{\lceil x \rceil - 1} = 2^{\lfloor x \rfloor - 1} \cdot ( 1 + 2 ) = 3 \cdot 2^{\lfloor x \rfloor - 1}. $$ So the exponent is reduced by one, e.g. $$ 2^{4.3} \approx 3 \cdot 2^3 = 24$$ Of course this trick aims to deal with decimal exponents.

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  • $\begingroup$ Down this same line would $(1+x-\lfloor x \rfloor)2^{\lfloor x \rfloor}$ work as well? This is off the cuff but it would be just as easy to calculate in one's head and may be a little closer to the actual answer. $\endgroup$ – Thomas Rasberry Apr 8 '17 at 19:03
  • $\begingroup$ @ThomasRasberry, interesting but how did you get to your expression? $\endgroup$ – Fabian Schn. Apr 8 '17 at 20:10
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    $\begingroup$ This isn't very accurate $\endgroup$ – David Quinn Apr 8 '17 at 20:29
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use taylor approximation , for instance

$2^{4.3} $

use integer part $a=4$ and fraction part $b=0.3$

now let us consider taylor approximation because, you have such kind of equation

$2^{a+b} $ which is the same as $f(x+\alpha)$ if $f(x)=2^x$

$$f(x) \approx f(a) + f'(a)(x-a)$$

derivative of $2^x=2^x\log(2)$

so we would have

$2^{4.3}=2^4+2^4*log(2)*(4.3-4)$

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    $\begingroup$ How would you calculate $\log(2)$ without calculator? $\endgroup$ – kingW3 Apr 8 '17 at 18:59
  • $\begingroup$ @kingW3 nice question :D :D :D i would leave it $\endgroup$ – dato datuashvili Apr 8 '17 at 19:00
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    $\begingroup$ @kingW3: It helps to know some constants if you want to do things without a calculator. $\log_{10} 2 \approx 0.30103$ and $\ln 2 \approx 0.693$ are early candidates. $\endgroup$ – Ross Millikan Apr 8 '17 at 19:51
  • $\begingroup$ thanks in advance @RossMillikan $\endgroup$ – dato datuashvili Apr 8 '17 at 19:52
  • $\begingroup$ @kingW3 The natural log of $2$ is about 0.7. That's the basis for the rule of 70 (en.wikipedia.org/wiki/Rule_of_72https://en.wikipedia.org/wiki/…) In Heinlein's Time for the Stars that's something the protagonist just knows: "'Point six nine three one" "That's the natural log of two." "What did you think it was?" page 12: books.google.com/…: $\endgroup$ – Ethan Bolker Apr 8 '17 at 20:58
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Memorize a few trick. $2^{10} = 1024\approx 10^3$ so $2^{3.3} \approx 10$.

So $2^{k + \frac i/3} = 2^{k-3i + i*3.3}= 2^{k-3i}10^i$ so $2^{4.3} = 2^{1 + 3.3} = 20$ ish.

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If you are happy to quote, without refernce to a calculator, that the approximate value of $\sqrt{2}$ is $1.4$ then, $$2^{4.3}=2^4\times2^{0.3}=16\times(\sqrt{2})^{0.6}\simeq16\times(1.4)^{0.6}$$ Now use the Binomial Theorem and get $$2^{4.3}\simeq 16\times(1+0.4\times0.6)=16\times1.24=19.84$$

This has an error of approximately 3%.

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