What is the approx. value of W, if W=(1.5)*11,Given log2=0.301, log 3=.477.
A) 68 B) 86 C) 105 D) 125

• if W =(1.5)*11 then simply,W = 16.5
  
  
  if, W=(1.5)^11
  take log both sides
  => logW = 11*log(1.5) =11*log(3/2)
  => logW = 11(log3-log2)= 11*0.176 = 1.936
  => W = 10^(1.936)
  => W = 86 (approax.)
• 10 years agoHelpfull: Yes(51) No(0)
• can anyone tell me how 10^(1.93) is calculated????????????
• 10 years agoHelpfull: Yes(9) No(0)
• Ans: 86
  
  Log W=Log(1.5^11)=11*Log(3/2)=11*( Log(3)-Log(2) )
  Log W=1.936
  W=10^1.936
  We know 10^2=100.
  So 10^1.936 will be close to and less than 100.
  So by options you can get the ans as 86)
  
  
• 8 years agoHelpfull: Yes(3) No(0)
• w=1.5^11
  logw=(log1-log2+log3)11
  logw=(0-.301+.477)11
  =(.176)11
  =1.986
  w=10^1.986
  =86(apprx)
• 10 years agoHelpfull: Yes(2) No(0)
• 86
  take log both sides, 1.5 is 3/2
  than, W=10^(log3-log2)
  
• 10 years agoHelpfull: Yes(1) No(0)
• hey, please explain the skyscraper problem in 8th june paper
• 10 years agoHelpfull: Yes(1) No(1)
• log(base 10)w = log(3/2)*11= 1.93
  w= 10^(1.93)= 86 (approx) [lof base(b)a =x
  then a=b^x ]
• 10 years agoHelpfull: Yes(1) No(0)
• Given that W=(1.5)^11 ,log2=.301,log=.477
  Taking common log on both sides,
  logW=log(1.5)^11
  logW=11*log(1.5) since log(x) ^a= a*log(x)
  logW=11*log(3/2) => 11*(log(3)-log(2))
  since log(x/y)=logx-logy
  logW=11*(.477-.301)=1.936
  Taking antilog on both sides,
  W=86.3 (approx).
• 5 years agoHelpfull: Yes(0) No(0)