approximation for value of $2^x$ without using calculator How to find an approximation for value of $2^x$ without using calculator?
For example, $2^{4.3}$. 
 A: 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.
A: 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
$2^{4.3}=2^4+2^4*log(2)*(4.3-4)$
A: 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.
A: 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%.
