If a number x^100 have 31 digit then how many digits are there in x^1000?

• x^100 have 31 digits => x = 2 only as 3^100 have more than 31.
  check, if
  y = 2^100
  => logy = 100*log2
  => logy = 100*.301
  => logy = 30.1
  => y has 30+1= 31 digits.
  
  so, x^1000 = 2^1000 = y
  => logy = 1000*log2
  => logy = 301
  => y has 301+1 = 302 digits.
  
• 10 years agoHelpfull: Yes(21) No(2)
• 10^1 has 2 digits = 10
  10^2 has 3 digits = 100
  10^3 has 4 digits = 1000
  similarly 10^y-1 has y digits
  if we have y digits in x^100 that means
  10^(y-1)=x^100
  10^(31-1)=x^100
  taking log both side
  30 log 10 = 100 log x
  30 = 100 log x
  now,
  10^(y-1)= x^1000
  y-1=1000 log x
  y-1=10*100 log x
  put value of 100 log x in this equation
  we get y-1=10*30
  y=301
  Ans=301
• 10 years agoHelpfull: Yes(4) No(4)
• y=x^100 [where y is 31]
  taking log both side
  log y = 100 log x
  log y = 100 log(0.3010) [assume the value of x]
  log y = 30.1 = 31 digit [value of x = 2 ]
  then
  y = 2 ^ 1000
  taking log both side
  log y = 1000 log 2
  log y = 1000*0.3010 = 301 (ans)
• 10 years agoHelpfull: Yes(3) No(0)
• i think 310
  
• 10 years agoHelpfull: Yes(2) No(1)
• I cant understand your question can u give me option ,,,,
• 10 years agoHelpfull: Yes(1) No(0)
• @VAMI : This ans wasn't thr in the option.
• 10 years agoHelpfull: Yes(0) No(0)
• Please enter the options as well..!!
  
• 10 years agoHelpfull: Yes(0) No(0)