@Dev Sharma
–
Actually I got the answer but according to my book answer key I was wrong so....Thanks
–
Naitik Sanghavi
·
1 year, 8 months ago

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@Dev Sharma
–
@Dev Sharma DevBut how will you find the number of digits without computing \(\large2^{22}\) if it is asked in exam?
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Naitik Sanghavi
·
1 year, 8 months ago

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@Naitik Sanghavi
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U may assume the no. Of digits of a number be x. Let the number be a^b. So
10^(x-1) = a^b
X-1 = blog(a) [log with base 10]
X = blog(a) +1
For convenience, X = [blog(a)] +1
So 2^22 has [22log(2)] +1 digits.similarly following can be calculat ed
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Aditya Narayan Sharma
·
1 year, 8 months ago

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@Aditya Narayan Sharma
–
How to apply here .here it will become 2^22 log(2) then we have to apply log again then do antilog.after doing this I got answer 447 .is it right??????
–
Anshul Sanghi
·
1 year, 6 months ago

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TopNewest1262612 – Dev Sharma · 1 year, 8 months ago

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– Naitik Sanghavi · 1 year, 8 months ago

Actually I got the answer but according to my book answer key I was wrong so....ThanksLog in to reply

@Dev Sharma DevBut how will you find the number of digits without computing

\(\large2^{22}\)if it is asked in exam? – Naitik Sanghavi · 1 year, 8 months agoLog in to reply

– Aditya Narayan Sharma · 1 year, 8 months ago

U may assume the no. Of digits of a number be x. Let the number be a^b. So 10^(x-1) = a^b X-1 = blog(a) [log with base 10] X = blog(a) +1 For convenience, X = [blog(a)] +1 So 2^22 has [22log(2)] +1 digits.similarly following can be calculat edLog in to reply

– Anshul Sanghi · 1 year, 6 months ago

How to apply here .here it will become 2^22 log(2) then we have to apply log again then do antilog.after doing this I got answer 447 .is it right??????Log in to reply

@Calvin Lin Please help me with this! – Naitik Sanghavi · 1 year, 8 months ago

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