## Yahoo! 3 years ago if a,b,c are in G.P then loga (x) , logb (x) , logc (x) are in ?

1. shaik0124

ap

2. Yahoo!

Nop ! @shaik0124

3. shaik0124

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!

But My book says it is H.P

5. shubhamsrg

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

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

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

8. sauravshakya

Thus they are in A.P

9. shubhamsrg

exactly!! as @sauravshakya said/did..

10. waterineyes

They will be in Harmonic Progression..

11. Yahoo!

Yup....they should be in HP ... But hw

12. Yahoo!

@waterineyes

13. shubhamsrg

ohh,,didnt i explain well @Yahoo! ?

14. waterineyes

Let me write.. It will take time.

15. Yahoo!

@shubhamsrg finally u agreed that it is Ap

16. sauravshakya

so 1/log a , 1/log b and 1/log c are in HP

17. shubhamsrg

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

18. shubhamsrg

which ultimately means the ans is HP as you are looking for 1/log a,1/log b and 1/log c

19. waterineyes

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

Now given: $b^2 = ac$

21. waterineyes

$\huge x^{\frac{2}{m}} = x^{\frac{1}{k}} \times x^{\frac{1}{n}}$

22. waterineyes

$\implies \frac{1}{k} + \frac{1}{n} = \frac{2}{m}$ This implies k, m and n are in HP..

23. waterineyes

How substitute back for k, m and n back..

24. waterineyes

*now

25. waterineyes

$k = \log_a(x), m = \log_b(x), n = \log_c(x)$

26. waterineyes

Getting @Yahoo!

27. Yahoo!

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

This you can do if you want to verify..

29. Yahoo!

i did by this and got answer as it is AP

30. waterineyes

Let me check then..

31. sauravshakya

can I help?

32. Yahoo!

Yup

33. sauravshakya

|dw:1344689783197:dw| right?

34. Yahoo!

Yup.. Base changing rule

35. waterineyes

They are in HP @Yahoo!

36. Yahoo!

hw @waterineyes

37. waterineyes

Let a = 2, b= 4 and c = 8 Let x = 64

38. Yahoo!

ok

39. sauravshakya

|dw:1344689862033:dw|similarly....

40. sauravshakya

right?

41. sauravshakya

@Yahoo!

42. waterineyes

$\log_2(64) = 6$ $\log_4(64) = 3$ $\log_8(64) = 2$ $\huge 6, 3, 2 \; \; are \; \; In \; \; HP..$

43. Yahoo!

if u take x =8

44. waterineyes

$\frac{2}{3} = \frac{1}{6} + \frac{1}{2}$

45. waterineyes

There is no condition you can take x = 8.. But the fact is that by taking 8 it will not become AP..

46. Yahoo!

log2 (8( = 4 log4(8) = 2 log 8 (8) = 1

47. waterineyes

$\huge \log_4(8) \ne 2$

48. Yahoo!

2^3/2^2 = 2

49. waterineyes

NO..

50. sauravshakya

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

$\huge \log_4(8) =\frac{3}{2}$

52. waterineyes

Use the base change property that @sauravshakya gave you..

53. Yahoo!

oh..ok.... i really need to study logarithms

54. Yahoo!

@waterineyes @sauravshakya @shubhamsrg thxxxxxx

55. waterineyes

Welcome dear..