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abdul_shabeer

  • 4 years ago

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

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  1. myininaya
    • 4 years ago
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    so you want that in terms of a and b and c?

  2. myininaya
    • 4 years ago
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    and z?

  3. abdul_shabeer
    • 4 years ago
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    In terms of y

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

  5. FoolForMath
    • 4 years ago
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    2/y

  6. abdul_shabeer
    • 4 years ago
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    Can you explain?

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

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

  9. myininaya
    • 4 years ago
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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?

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

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

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

  13. abdul_shabeer
    • 4 years ago
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    No solution

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

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

  16. FoolForMath
    • 4 years ago
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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 ;)

  17. abdul_shabeer
    • 4 years ago
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    Thanks ffm

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  34. FoolForMath
    • 4 years ago
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    Thanks myin :)

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