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8 lessons,
56m 1s
Practice Questions : Part 4
5 plays

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.

Unacademy user
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


  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.