anonymous
  • anonymous
Given that 2.004
Algebra
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
10101 in binary? then 210 shouldn't be binary....
anonymous
  • anonymous
( 101101)base2=( 45)decimal binary no always power of 2 therefore 2power 6= 64 hence 6 digit req..
anonymous
  • anonymous
brilliant!

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anonymous
  • anonymous
it is 101^101
anonymous
  • anonymous
@ParthKohli
anonymous
  • anonymous
help!
ParthKohli
  • ParthKohli
Well, find out \(\log_{10} 101^{101}\)
anonymous
  • anonymous
then
ParthKohli
  • ParthKohli
That's it.
anonymous
  • anonymous
that is the no. of digits?
ParthKohli
  • ParthKohli
Yes
anonymous
  • anonymous
how log can help calculate no. of digits?
ParthKohli
  • ParthKohli
Yes, it does. Try it!
anonymous
  • anonymous
202 is incorrect
ParthKohli
  • ParthKohli
You also have to consider the assumptions given in your question.
anonymous
  • anonymous
what else is given?
ParthKohli
  • ParthKohli
Actually, you have to find \(1 + \log_{10} 101^{101}\). Don't write in the answer, let me check first.
anonymous
  • anonymous
ok
ParthKohli
  • ParthKohli
You can do something else while I check. Thanks :-)
anonymous
  • anonymous
ok
ParthKohli
  • ParthKohli
203.
ParthKohli
  • ParthKohli
And so I was correct :-)
anonymous
  • anonymous
why add 1?
ParthKohli
  • ParthKohli
Because the number of digits in \(100\) is not \(\log_{10} 100\), but it's \(1+\log_{10}100\)
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
can it be used for counting digits in any expansion?
ParthKohli
  • ParthKohli
Expansion? You mean number base?
anonymous
  • anonymous
like this type of question
ParthKohli
  • ParthKohli
Yes.
anonymous
  • anonymous
ok thanks

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