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?


EDIT: Sorry, I think I can make this clearer, how can I solve:

$2^x = 10^{80}$


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 $\log 10 = 1$)


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, $\log_{10} 2 \approx 0.3$.)


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: $1234 = y^e$.

  • $\begingroup$ Thank you! Sorry for the long winded question, this will work fine. I'll accept as soon as it lets me. $\endgroup$ – user2200321 Nov 22 '13 at 23:22
  • $\begingroup$ You're welcome ! $\endgroup$ – Sudarsan Nov 23 '13 at 0:24

$10^y = 2^x , x=y/\log(2)$


$2^x = 10^{x\log(2)}$

Hopefully this helps explain:

$2^x = 10^{x\log(2)}$

$\log(2^x) = \log(10^{x\log(2)})$

$x\log(2) = x\log(2)\log(10)$

$\log(2) = \log(2)\log(10)$

$\frac{\log(2)}{\log(2)} = \log(10)$


  • $\begingroup$ you assumed the consequent in your proof... $\endgroup$ – louie mcconnell Nov 6 '19 at 11:50
  • $\begingroup$ I guess it's not a proof, it helped me grasp the concept though. $\endgroup$ – JeremiahBarrar Nov 20 '19 at 3:10

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