Convert from high exponent of base $10$ to base $2$. Is there an efficient way to convert from a high exponent of base $10$, to base $2$? Both in exponent notation. Here's an example:
If I have a number that's $10^5$ or even $10^{100}$, and I wanted to convert that to base $2$, exponent notation, how could I do that? Normally, I would do:
$10^5 = 100000$, then go back and convert it to base $2$, but can I just an operation to the exponent form of $10^5$, and find the answer?
Thanks!
EDIT: Sorry, I think I can make this clearer, how can I solve:
$2^x = 10^{80}$
 A: Rough rule of thumb: $2^{10} \approx 10^3$.  So if $2^x = 10^{80}$, then $x \approx \frac{10}3 \times 80$.  (Or to line up with the other answers, $\log_{10} 2 \approx 0.3$.)
A: Take the logarithm of the base that you want with respect to the original base. For example, to convert from base $x$ of a number, say $1234$, to base $y$, do the following: $\frac{\log_x(1234)}{\log_y(1234)}$. The result gives you the exponent $e$ to which $y$ has to be raised to get the number $1234$. Result: $1234 = y^e$.
A: You could also solve it like this. 
Log both sides.
$\log 2^x = \log 10^{80}$
Using $\log a^r = r \cdot \log a$ (a rule of logarithms) the expression above becomes
$x \cdot \log 2 = 80 \cdot \log 10$
$\log 2$ & $\log 10$ are simply constants that you can plug into a calculator, so the expression becomes
$x = \frac{80\cdot \log 10}{\log2} \implies x = \frac{80}{\log 2}$
(since $\log 10 = 1$)
A: $10^y = 2^x , x=y/\log(2)$
Also:
$2^x = 10^{x\log(2)}$
Hopefully this helps explain:
$2^x = 10^{x\log(2)}$
$\log(2^x) = \log(10^{x\log(2)})$
$x\log(2) = x\log(2)\log(10)$
$\log(2) = \log(2)\log(10)$
$\frac{\log(2)}{\log(2)} = \log(10)$
$1=1$
