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10101 in binary? then 210 shouldn't be binary....

( 101101)base2=( 45)decimal binary no always power of 2 therefore 2power 6= 64
hence 6 digit req..

brilliant!

it is 101^101

help!

Well, find out \(\log_{10} 101^{101}\)

then

That's it.

that is the no. of digits?

Yes

how log can help calculate no. of digits?

Yes, it does. Try it!

202 is incorrect

You also have to consider the assumptions given in your question.

what else is given?

ok

You can do something else while I check. Thanks :-)

ok

203.

And so I was correct :-)

why add 1?

Because the number of digits in \(100\) is not \(\log_{10} 100\), but it's \(1+\log_{10}100\)

ok thanks

can it be used for counting digits in any expansion?

Expansion? You mean number base?

like this type of question

Yes.

ok thanks