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King

  • 4 years ago

Its a log problem which has sqrt so i shall post in reply...

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  1. King
    • 4 years ago
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    \[\log_{4} \sqrt[4]{32\sqrt[3]{16\sqrt{64.8^{-3\div4}}}}\]

  2. Ishaan94
    • 4 years ago
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    2^5 = 32 2^4 = 16 Is it 64 x 8? or 64.8

  3. King
    • 4 years ago
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    its 64 x 8^-3/4

  4. King
    • 4 years ago
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    please give full explanation

  5. Ishaan94
    • 4 years ago
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    2^3 = 8 2^6 = 64 Now try it

  6. Ishaan94
    • 4 years ago
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    I don't like this problem

  7. King
    • 4 years ago
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    wait even i dont like it

  8. Ishaan94
    • 4 years ago
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    It's not hard but it's unnecessary

  9. King
    • 4 years ago
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    its necessary for me

  10. sheg
    • 4 years ago
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    1.019

  11. King
    • 4 years ago
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    i am not getting it can u xplain

  12. Ishaan94
    • 4 years ago
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    oh cool sheg

  13. King
    • 4 years ago
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    no sheg answer is wrong

  14. sheg
    • 4 years ago
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    how come plz explain m right

  15. King
    • 4 years ago
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    idk answer given as 121/186 or 0.6505376

  16. sheg
    • 4 years ago
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    ok wait i made one mistake now i got it

  17. Ishaan94
    • 4 years ago
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    I am getting .7 after simplifying and calculating it in the calculator

  18. King
    • 4 years ago
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    do not use claculator

  19. King
    • 4 years ago
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    and explain

  20. King
    • 4 years ago
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    u should get it as 121/186

  21. Ishaan94
    • 4 years ago
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    King It's a big waste of time, it's not worth it

  22. King
    • 4 years ago
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    pls ishaan its necessary fr me

  23. King
    • 4 years ago
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    sheg wat abt u?

  24. sheg
    • 4 years ago
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    trying to solve it manually

  25. Ishaan94
    • 4 years ago
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    Hmm I am not posting the complete solution, Just a little simplification \[\frac{1}{4} \left(\log_4 32 + \frac{1}{3}\left(\log_416 + \frac{1}{2}\left(\log_4 64 + \log_4 8^{\frac{-3}{4}} \right)\right)\right)\]

  26. sheg
    • 4 years ago
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    is that 64 times 8?????????????

  27. Ishaan94
    • 4 years ago
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    Now you should do it on your own, I really don't like calculating Numbers

  28. King
    • 4 years ago
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    how did u get 1/4

  29. King
    • 4 years ago
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    yeah sheg

  30. sheg
    • 4 years ago
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    oh got i was taking 64.8

  31. Ishaan94
    • 4 years ago
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    Yeah Sheg

  32. Ishaan94
    • 4 years ago
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    Read the FIITJEE Module for Logs it's the basic property of Log

  33. sheg
    • 4 years ago
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    nahi yar mein usko decimal mein le raha tha na bracket use kiya hai isne na kuch

  34. King
    • 4 years ago
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    wait im reading and i gave u this sum from FIITJEE compendium

  35. King
    • 4 years ago
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    sry sheg are u now getting right answer?

  36. Ishaan94
    • 4 years ago
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    Hmm mujhe bhi same confusion hua tha

  37. sheg
    • 4 years ago
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    i m getting this thing \[\log_{4} 2^{\frac{167}{96}}\]

  38. King
    • 4 years ago
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    what is that 2^?

  39. sheg
    • 4 years ago
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    167/96

  40. King
    • 4 years ago
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    thats 0.8 sumthing...

  41. King
    • 4 years ago
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    ishaan i calculated ure equation i got 29/32

  42. Ishaan94
    • 4 years ago
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    you must have calculated it wrong

  43. King
    • 4 years ago
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    wait i am checking for mistake

  44. Ishaan94
    • 4 years ago
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    I am getting 127/192 I hate this problem, I hate calculating numbers

  45. Ishaan94
    • 4 years ago
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    .661458333 <<Calculator

  46. King
    • 4 years ago
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    u need to subtract 6 frm numerator and denom. to get rite answer so thatt means u are very close

  47. KatrinaKaif
    • 4 years ago
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    Took me forever to get to the bottom of this >.>

  48. King
    • 4 years ago
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    so post solution katrina

  49. King
    • 4 years ago
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    ishaan what is simplified version of log8^-3/4 base4

  50. Tomas.A
    • 4 years ago
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    \[\log_4\sqrt[4]{32\sqrt[3]{16\sqrt{64\cdot8^{-\frac{3}{4}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\sqrt{2^6\cdot2^{-\frac{9}{4}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\sqrt{2^{\frac{15}{4}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{2^4\cdot{2^{\frac{15}{8}}}}}=\] \[\log_4\sqrt[4]{2^5\sqrt[3]{{2^{\frac{47}{8}}}}}=\] \[\log_4\sqrt[4]{2^5\cdot{{2^{\frac{47}{24}}}}}=\] \[\log_4\sqrt[4]{{{2^{\frac{167}{24}}}}}=\] \[\log_{2^2}2^{\frac{167}{96}}=\] \[\log_{2}2^{\frac{167}{192}}=\] \[\frac{167}{192}\]

  51. Ishaan94
    • 4 years ago
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    Yeah I did it again I am getting the same

  52. King
    • 4 years ago
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    can u explain the 4th step

  53. King
    • 4 years ago
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    how did u take out 2^15/4?

  54. Tomas.A
    • 4 years ago
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    \[2^4\cdot2^{\frac{15}{8}}=2^{4+\frac{15}{8}}=2^{\frac{32}{8}+\frac{15}{8}}=2^{\frac{47}{8}}\] this one?

  55. King
    • 4 years ago
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    no how did u take 2^15/4 out frm sqrt

  56. Ishaan94
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=log_4%28%2832+*+%2816+*%2864+*+8^%28-3%2F4%29%29^%281%2F2%29%29^%281%2F3%29%29^%281%2F4%29 Tomas Is Right, FIITJEE is wrong!

  57. King
    • 4 years ago
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    ok now can u xplain

  58. Tomas.A
    • 4 years ago
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    \[\sqrt{2^\frac{15}{4}}=2^{\frac{15}{4}\cdot\frac{1}{2}}=2^\frac{15}{8}\]

  59. King
    • 4 years ago
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    oh right

  60. Tomas.A
    • 4 years ago
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    \[\sqrt[n]{m}=m^{\frac{1}{n}}\]

  61. King
    • 4 years ago
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    Yipppeeeeeeeee u guys are great!!!!!!!!!! thnx to all of u.!!!!!!!!!!!!!!!

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