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  • Question
  • The value of log 9/8 - log 27/32 + log3/4 is?


  • Options
  • A. 0
  • B. 1
  • C. 2
  • D. 3

  • Correct Answer


  • Explanation

    Given Exp. = log [{(9/8) / (27/32)} x 3/4)]
    = log [(9/8) x (3/4) x (32/27)]
    = log 1
    = 0


  • Logarithm problems


    Search Results


    • 1. 
      Given that log 10 2 = 0.3010, then log 2 10 is equal to?

    • Options
    • A. 0.3010
    • B. 0.6990
    • C. 1000 / 301
    • D. 699 / 301
    • Discuss
    • 2. 
      The value of log 23 x log 32 x log 34 x log 43 is?

    • Options
    • A. 1
    • B. 2
    • C. 3
    • D. 4
    • Discuss
    • 3. 
      The value of $ \frac{\log_{a}{x}}{\log_{ab}{x}} - \log_{a}{b}$ is?

    • Options
    • A. 0
    • B. 1
    • C. a
    • D. ab
    • Discuss
    • 4. 
      If log 2 = 0.3010, then the number of digits in 2 64 is?

    • Options
    • A. 18
    • B. 19
    • C. 20
    • D. 21
    • Discuss
    • 5. 
      Two poles of ht. a and b meters are p meters apart ( b > a). The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is :

    • Options
    • A. a+b ab
    • B. p a+b
    • C. ab a+b
    • D. a+b p
    • Discuss
    • 6. 
      The simplified form of log(75/16) -2 log(5/9) +log(32/343) is?

    • Options
    • A. log 2
    • B. 2 log 2
    • C. log 3
    • D. log 5
    • Discuss
    • 7. 
      If log 2 = 0.3010 then log 5 equals to?

    • Options
    • A. 0.3010
    • B. 0.6990
    • C. 0.7525
    • D. Given log 2, it is not possible to calculate log 5
    • Discuss
    • 8. 
      If log 102 =0.3010 and log 107 = 0.8451, then the value of log 10 2.8 is?

    • Options
    • A. 0.4471
    • B. 1.4471
    • C. 2.4471
    • D. 14.471
    • Discuss
    • 9. 
      If log 10 2 =0.301, then the value of log 10(50) is?

    • Options
    • A. 0.699
    • B. 1.301
    • C. 1.699
    • D. 2.301
    • Discuss
    • 10. 
      Find the value of log (a 2 / bc) + log (b 2 / ac) + log (c 2 / ab)?

    • Options
    • A. 0
    • B. 1
    • C. abc
    • D. ab2c2
    • Discuss


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