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Practice Questions : Part 4
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This lesson will help you answer the questions based on logarithms

Abhishek Khurana
I am Abhishek Khurana. I completed my B.Com(H) from Panjab University in 2008 & my MBA from NMIMS in 2011.

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How the Sample size is taken? Is it depend on Population?
Thera are various formulae, and apart from population, accuracy is also considered and other parameters too
  1. Abhishek Khurana 6 times 99 percentiler in CAT MBA from NMIMS 2009-2011 8 years of teaching experience


  2. LOGARITHMS


  3. Miscellaneous Examples Example 20: If log(2-1)(x-1)I = 1, then how many values can x take? Solution: lloga-)x-1)1 1 where x-1 >0 and 2x-1#1 (as logab is defined only when b>0 and a #1) Also log a bay=> a-b Now log(2-1)(x-1) =1 or log a-)/x-1)-1 case 1 : log(2-1) (x-1) = 1 2x-13x-1 =>x=0 not possible as x>1


  4. Miscellaneous Examples Example 21: Minimum value for the expression: 4logiox-logx (1/1000) where x>1 is a. 8 C. 4V3 d. 7 Solution: 4loghox-log (1/1000) 4log1o x 3/logiox using concept of AM > GM, 410g10 x + 3/log,ox/2 >s( 4logo x" 3/logeX


  5. Miscellaneous Examples Example 22: log(mtn3),log(m3 n3 )log(m2n2 )are the first three terms of an Arithmetic Progression. If the 8th term of the Arithmetic Progression is log(m n ), what is the value of a? Solution: log A, log B and log C are in AP, then A, B, and C are in G.P where BA2 AC. n=m The 8th term if the series would be n . So, the value of a = 0.


  6. Miscellaneous Examples Example 23: If log 0.125(0.25) X. log (41/3) 4 + log 4343T, then find the value of a. 343 b. 1/343 c. 1/49 cf. 1/7 Solution: log a.125(0.25) log (41/3)4+log 34317 2/3-3+log 77


  7. Miscellaneous Examples Example 24: If log2 x = logoY = m, and x and y are positive integers, then find logx Y- Solution: log2 X-m 1 /3 log-Y-m logx y -log2 y/log2 x 3m/m-3


  8. Miscellaneous Examples Example 25: If p2 x Q3 = 1000 and P2n x Qs, = 100 Ren, then find the value of log1o R. Solution: P2x Q3-1000 10A3 (paxQ3 ) n# 102 Ren Taking log on both sides (3n-2) log 10-6n log R 3n-2-6n log R log R-1/2-1/3n


  9. Miscellaneous Examples Example 26: If 2logx+41ogy+ 610gs122 1. 2, then the minimum possible value of (x2y+yz+ z2x) is Solution: loga logsy log22 logs(xyz)#2 2 (xyz22 64 (ryz) xyz 28 (xyz)s 2 512


  10. Miscellaneous Examples (x2 y). (ya z) (za x )28.8.8. Applying AM a GM we get that xy+yz+ 23 xy yazz2x)13 xy yz+2x 2 24 Minimum possible valued (x2y +y2z+ z2x) 24.