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Yahoo!

  • 3 years ago

if a,b,c are in G.P then loga (x) , logb (x) , logc (x) are in ?

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  1. shaik0124
    • 3 years ago
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    ap

  2. Yahoo!
    • 3 years ago
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    Nop ! @shaik0124

  3. shaik0124
    • 3 years ago
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    If a, b, and c are in GP, then b = ak and c = ak^2 for some constant k. This gives: log₂(b) = log₂(ak) = log₂(a) + log₂(k) log₂(c) = log₂(ak^2) = log₂(a) + 2log₂(k). This shows that log₂(a), log₂(b), and log₂(c) form a AP with a first term of log₂(a) and a common difference of log₂(k). Therefore, the answer is a) in AP. I hope this helps!

  4. Yahoo!
    • 3 years ago
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    But My book says it is H.P

  5. shubhamsrg
    • 3 years ago
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    we have ac = b^2 and we need to find about 1/loga , 1/log b , 1/log c where all bases are converted to x now clearly, log a + log c = log(ac) = 2logb thus log a,log b and log c are in AP so 1/log a , 1/log b and 1/log c are in HP

  6. sauravshakya
    • 3 years ago
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    SINCE they are in G.P b^2 = ac or, logac/logb = 2 or, loga + log c = 2 logb now, logb= loga+logc /2

  7. shubhamsrg
    • 3 years ago
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    the 3rd last line can also be understood if you take log to base x on both sides of ac= b^2

  8. sauravshakya
    • 3 years ago
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    Thus they are in A.P

  9. shubhamsrg
    • 3 years ago
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    exactly!! as @sauravshakya said/did..

  10. waterineyes
    • 3 years ago
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    They will be in Harmonic Progression..

  11. Yahoo!
    • 3 years ago
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    Yup....they should be in HP ... But hw

  12. Yahoo!
    • 3 years ago
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    @waterineyes

  13. shubhamsrg
    • 3 years ago
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    ohh,,didnt i explain well @Yahoo! ?

  14. waterineyes
    • 3 years ago
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    Let me write.. It will take time.

  15. Yahoo!
    • 3 years ago
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    @shubhamsrg finally u agreed that it is Ap

  16. sauravshakya
    • 3 years ago
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    so 1/log a , 1/log b and 1/log c are in HP

  17. shubhamsrg
    • 3 years ago
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    lol..noo..i didnt.. i meant you can derive loga ,log b and log c are in AP but @sauravshakya 's method..

  18. shubhamsrg
    • 3 years ago
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    which ultimately means the ans is HP as you are looking for 1/log a,1/log b and 1/log c

  19. waterineyes
    • 3 years ago
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    Let: \[\large \log_a(x) = k \implies a^k = x \implies a = x^{\frac{1}{k}}\] Similarly: Let others: \[\large b = x^{\frac{1}{m}}\] \[\large c = x^{\frac{1}{n}}\]

  20. waterineyes
    • 3 years ago
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    Now given: \[b^2 = ac\]

  21. waterineyes
    • 3 years ago
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    \[\huge x^{\frac{2}{m}} = x^{\frac{1}{k}} \times x^{\frac{1}{n}}\]

  22. waterineyes
    • 3 years ago
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    \[\implies \frac{1}{k} + \frac{1}{n} = \frac{2}{m}\] This implies k, m and n are in HP..

  23. waterineyes
    • 3 years ago
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    How substitute back for k, m and n back..

  24. waterineyes
    • 3 years ago
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    *now

  25. waterineyes
    • 3 years ago
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    \[k = \log_a(x), m = \log_b(x), n = \log_c(x)\]

  26. waterineyes
    • 3 years ago
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    Getting @Yahoo!

  27. Yahoo!
    • 3 years ago
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    Can we subs the value for a b c such as 2 , 4 , 8 and assume value for x as 8 and solve the log??

  28. waterineyes
    • 3 years ago
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    This you can do if you want to verify..

  29. Yahoo!
    • 3 years ago
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    i did by this and got answer as it is AP

  30. waterineyes
    • 3 years ago
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    Let me check then..

  31. sauravshakya
    • 3 years ago
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    can I help?

  32. Yahoo!
    • 3 years ago
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    Yup

  33. sauravshakya
    • 3 years ago
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    |dw:1344689783197:dw| right?

  34. Yahoo!
    • 3 years ago
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    Yup.. Base changing rule

  35. waterineyes
    • 3 years ago
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    They are in HP @Yahoo!

  36. Yahoo!
    • 3 years ago
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    hw @waterineyes

  37. waterineyes
    • 3 years ago
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    Let a = 2, b= 4 and c = 8 Let x = 64

  38. Yahoo!
    • 3 years ago
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    ok

  39. sauravshakya
    • 3 years ago
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    |dw:1344689862033:dw|similarly....

  40. sauravshakya
    • 3 years ago
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    right?

  41. sauravshakya
    • 3 years ago
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    @Yahoo!

  42. waterineyes
    • 3 years ago
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    \[\log_2(64) = 6\] \[\log_4(64) = 3\] \[\log_8(64) = 2\] \[\huge 6, 3, 2 \; \; are \; \; In \; \; HP..\]

  43. Yahoo!
    • 3 years ago
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    if u take x =8

  44. waterineyes
    • 3 years ago
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    \[\frac{2}{3} = \frac{1}{6} + \frac{1}{2}\]

  45. waterineyes
    • 3 years ago
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    There is no condition you can take x = 8.. But the fact is that by taking 8 it will not become AP..

  46. Yahoo!
    • 3 years ago
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    log2 (8( = 4 log4(8) = 2 log 8 (8) = 1

  47. waterineyes
    • 3 years ago
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    \[\huge \log_4(8) \ne 2\]

  48. Yahoo!
    • 3 years ago
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    2^3/2^2 = 2

  49. waterineyes
    • 3 years ago
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    NO..

  50. sauravshakya
    • 3 years ago
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    Now,SINCE they are in G.P b^2 = ac or, logac/logb = 2 or, loga + log c = 2 logb now, logb= loga+logc /2 Thus,1/log a , 1/log b and 1/log c are in HP So, logx/loga , logx/logb and logx/logc are in HP thus, loga (x) , logb (x) , logc (x) are in HP

  51. waterineyes
    • 3 years ago
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    \[\huge \log_4(8) =\frac{3}{2}\]

  52. waterineyes
    • 3 years ago
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    Use the base change property that @sauravshakya gave you..

  53. Yahoo!
    • 3 years ago
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    oh..ok.... i really need to study logarithms

  54. Yahoo!
    • 3 years ago
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    @waterineyes @sauravshakya @shubhamsrg thxxxxxx

  55. waterineyes
    • 3 years ago
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    Welcome dear..

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