How to convert from a power of base two to a power of base 10? I might have an extremely silly question:
If I have a number, say $2^{32}$ and I need to convert to base 10, how should I do it? I know it should be $4 * 10^9$, but I do not know how did we get it.
I understand that $10$ is $2^3 + 2$, but I cannot understand how to proceed further in my reasoning...
Thanks!
 A: To brush up on algebra rules for exponents, the relevant rule here is that $x^{A*B}=(x^A)^B$. And as the others have already mentioned, $2^{10}=1024≈10^3=1000$. If we pick an example number that is so horrendously gigantic where this small inaccuracy doesn't matter, something like $2^{(2^{20})}=2^{1,048,576}$, you can factor out the exponent and replace it as follows:
$$2^{1,048,576}=2^{10*104,857}=(2^{10})^{104,857}=(10^3)^{104,857}=10^{3*104,857}=10^{314,571}$$
Extremely simple, as long as you remember the algebra fundamentals :)
A: To work with exponents of different bases, you also need its opposite operation, the logarithm:
$$
2^x = 10^y
$$
applying a $log_{10}$ to the previous equation:
$$
y = \log_{10} 2^x = x·\log_{10}2 = 0.30103x
$$
Therefore (in your question; $2^{32}$):
$$
y = 32×0.30103 = 9.63296
$$
$$
2^{32} \sim 10^{9.6}
$$
Summary:
The factor $0.3$ can be used to multiply the exponent of $2$ to obtain the exponent of $10$, or can be used to divide the exponent of $10$ to get the exponent of $2$, and the approximation is very high, particularly for big exponents.
A: Short answer
$2^{10} \approx 10^3$
Explanation
$$ \log_{10} (2^{32}) = \frac{32 \log 2}{\log 10} \approx 9.633
$$
and since $10^{0.633} \approx 4$,
$$ 2^{32} \approx 4 \cdot 10^9$$.
Or, notice that $2^{10} \approx 10^3$. It follows that $2^{30} \approx 10^9$, and $2^{32} \approx 4 \cdot 10^9$.
A: Let's start by noting that $2^{32}\ne 4*10^{9}$. However, $4*10^{9}$ does approximate $2^{32}$ to one significant figure ($2^{32}=4294967296$). So, strictly speaking, you do not seem to know what the answer should be---only a low-order approximation.
There is probably a sophisticated way of solving this problem algebraically... But we also have the option to simply brute-force it. 
By "brute forcing," I mean to just write out $2^{32}$ in its integer form by plugging it in to a calculator. This will give you its base 10 representation by default. You will find that you get:
$$2^{32}=4294967296$$
This number is actually short-hand notation for an expansion in powers of 10. In general, any integer can be written:
$$N = \Sigma_{i=0}^{M}A_{i}10^{i},\quad where\quad M \lt \infty$$
It is easiest to see this by starting from the right-side of 4294967296, which represents the single digit term ($6x10^{0}$). The number preceding the single-digit term, to the left, represents a multiple of 10 ($9x10^{1}$), while the third number from the right represents a multiple of 100 ($2x10^{2}$), and so on:
$$2^{32} = 4294967296 \\ =  6(10^{0})+9(10^{1})+2(10^{2})+7(10^{3})\\+6(10^{4})+9(10^{5})+4(10^{6})+9(10^{7})\\+2(10^{8})+4(10^{9})$$
From this expansion, you can see that $4*10^{9}$ is only the lowest order approximation. There are higher-order corrections all the way down to the single-digit scale! 
Of course, writing this expansion out explicitly is not necessary. The exact base-10 representation you are looking for is simply found by simultaneously multiplying and dividing the integer-representation we found by its leading power of 10 (which is equivalent to multiplying by 1):
$$2^{32}*\frac{10^{9}}{10^{9}}=\frac{4294967296}{10^{9}}*10^{9}=4.294967296*10^{9}$$
