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abdul_shabeer

If a^x=b^y=c^z and b^2=ac, then (1/x)+(1/y) is

  • 2 years ago
  • 2 years ago

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  1. myininaya
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    so you want that in terms of a and b and c?

    • 2 years ago
  2. myininaya
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    and z?

    • 2 years ago
  3. abdul_shabeer
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    In terms of y

    • 2 years ago
  4. abdul_shabeer
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    (a) 2/y (b) 1/y (c) 1/2y (d) 2y

    • 2 years ago
  5. FoolForMath
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    2/y

    • 2 years ago
  6. abdul_shabeer
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    Can you explain?

    • 2 years ago
  7. FoolForMath
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    Sure I can, tell me have you tried taking logarithms?

    • 2 years ago
  8. abdul_shabeer
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    No, I don't think this question should be solved by logarithms

    • 2 years ago
  9. myininaya
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    How do you think we should solve it? Are do you have any thoughts? Or is this a test for us?

    • 2 years ago
  10. abdul_shabeer
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    I had this doubt from a book, under the topic algebra.

    • 2 years ago
  11. myininaya
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    how did they do it?

    • 2 years ago
  12. myininaya
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    or they didn't?

    • 2 years ago
  13. abdul_shabeer
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    No solution

    • 2 years ago
  14. myininaya
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    i want to see ffm's explanation :)

    • 2 years ago
  15. vicky007
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    Is the question 1/x+1/z ? then it is 2/y

    • 2 years ago
  16. FoolForMath
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    Okay without logarithms, Set \( a^x = b^y = c^z = k \) Then, \( a = k^{1/x}, b = k^{1/y}, c = k^{1/z} \) Given \( b^2 = ac \). So, \( (k^{1/y} )^2 = \( k^{1/x} \times k^{1/z} \) \( \Rightarrow k^{2/y} = k^{ (1/x) + (1/z)} \) Hence, 2/y = (1/x) + (1/z) QED ;)

    • 2 years ago
  17. abdul_shabeer
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    Thanks ffm

    • 2 years ago
  18. TuringTest
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    we wanted 1/x+1/y though you have 1/x+1/z....?

    • 2 years ago
  19. FoolForMath
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    Glad to help :) and my apologies for the buggy Latex :(

    • 2 years ago
  20. TuringTest
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    am I missing something?

    • 2 years ago
  21. abdul_shabeer
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    Sorry, it's (1/x)+(1/z)

    • 2 years ago
  22. TuringTest
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    ah, then FFM has done it again :D

    • 2 years ago
  23. FoolForMath
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    lol, it's was pretty obvious, also it's a typical JEE problem :D

    • 2 years ago
  24. TuringTest
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    what's JEE ?

    • 2 years ago
  25. vicky007
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    @FFM are you an IIT aspirant?

    • 2 years ago
  26. vicky007
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    so now you are an iitian. I am writing the test this year

    • 2 years ago
  27. FoolForMath
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    Turing, all of this: http://en.wikipedia.org/wiki/JEE

    • 2 years ago
  28. FoolForMath
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    Except the last J2EE.

    • 2 years ago
  29. abdul_shabeer
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    This question was in a eigth grade book

    • 2 years ago
  30. FoolForMath
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    Lol, great :)

    • 2 years ago
  31. FoolForMath
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    That implies I am smarter than a 8 grader ? :P

    • 2 years ago
  32. myininaya
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    lol i thought it weird to have a y in the expression to start with

    • 2 years ago
  33. myininaya
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    nice job ffm :)

    • 2 years ago
  34. FoolForMath
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    Thanks myin :)

    • 2 years ago
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