Solve
100^x(log 9 base x - log root3 base x)
X is Greater then 1.
a)0,Infinity
b)-Infinity , Infinity
c)Infinity,-Infinity
d)None

• 100^x(log 9 base x - log root3 base x)
  
  100^x(2*log3 base x - (1/2) *log3 base x)
  
  100^x(3/2)*log3 base x
  
  as x > 1
  for any values of x, given exp is always +ve, it is never equal to zero or -Infinity.
  
  d)None
• 10 years agoHelpfull: Yes(23) No(0)
• Hello Rakesh If the Statement in always +ve Then We can say that the Value is in between 0 to infinity?????
• 10 years agoHelpfull: Yes(3) No(0)
• -infinity,0
  
• 10 years agoHelpfull: Yes(0) No(4)
• Not in the Option so u are wrong
• 10 years agoHelpfull: Yes(0) No(0)
• 100^x(log 3 base x-log 9 base root x) so when we simplify this we'll get -3*100^x/log x base 3 and in the question log x base 3 was greater than 1(log x base 3>1)so obviously x would be greater than 3(x>3) no any value u put u'll get the negative value only so option does not match and
  answer (d)
  
• 10 years agoHelpfull: Yes(0) No(2)
• 100^x(log 9 base x - log root3 base x)
  
  100^x(2*log3 base x - (1/2) *log3 base x)
  
  100^x(3/2)*log3 base x
  
  as x > 1
  for any values of x, given exp is always +v
  so the expression value will lie in between 0 and infinity..and the range does not include both the starting and end points here, so the option looks more appropriate.
  
• 10 years agoHelpfull: Yes(0) No(0)
• According to question is x>1 then it means
  the obtained equation 100^x3/2ln (base 3) x
  will be never 0 so option must be D.
• 10 years agoHelpfull: Yes(0) No(0)