If x^100 is 31 digit number then x^1000 contains how many digits.

• no=x^31
  no of digits in any no =floor[log(n)]+1
  so 100logx+1=31
  100logx=30
  hence
  1000logx+1=300+1=301
  so answer is 301
  
• 10 years agoHelpfull: Yes(26) No(12)
• 10^1 has 2 digits = 10
  10^2 has 3 digits = 100
  10^3 has 4 digits = 1000
  10^30 has 31 digits
  therefore, 10^30=x^100
  30=100 log x
  multiplaying both side by 10
  30*10=10*100 log x
  300= 1000 log x = 300+1=301
  so 301 is correct answer
• 10 years agoHelpfull: Yes(15) No(5)
• ans is 302
  log 2 x 1000=301.029995664
  choose higher natural no
• 9 years agoHelpfull: Yes(3) No(0)
• x^100 have 31 digits => x = 2 only as 3^100 have more than 31. but for x=2 no of digit will be 30
  so, x^1000 = 2^1000 = y
  => logy = 1000*log2=1000*0.301
  => logy = 301
  so y has 301+1 = 302 digits.
• 9 years agoHelpfull: Yes(2) No(0)
• if number is 2 digit it is 10 if number is 3 digit 10^2............if x digit then 10^(x-1)
  10^(x-1)=x^100 take log both side (x-1)=100 logx
  x-1=31
  logx=0.31
  similarly10^(a-1)= x^1000
  take log both side
  a-1=1000 logx
  put the value of logx
  a=311
• 10 years agoHelpfull: Yes(1) No(12)