anonymous
  • anonymous
if a,b,c are in G.P then loga (x) , logb (x) , logc (x) are in ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
shaik0124
  • shaik0124
ap
anonymous
  • anonymous
Nop ! @shaik0124
shaik0124
  • 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!

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anonymous
  • anonymous
But My book says it is H.P
shubhamsrg
  • 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
anonymous
  • anonymous
SINCE they are in G.P b^2 = ac or, logac/logb = 2 or, loga + log c = 2 logb now, logb= loga+logc /2
shubhamsrg
  • shubhamsrg
the 3rd last line can also be understood if you take log to base x on both sides of ac= b^2
anonymous
  • anonymous
Thus they are in A.P
shubhamsrg
  • shubhamsrg
exactly!! as @sauravshakya said/did..
anonymous
  • anonymous
They will be in Harmonic Progression..
anonymous
  • anonymous
Yup....they should be in HP ... But hw
anonymous
  • anonymous
@waterineyes
shubhamsrg
  • shubhamsrg
ohh,,didnt i explain well @Yahoo! ?
anonymous
  • anonymous
Let me write.. It will take time.
anonymous
  • anonymous
@shubhamsrg finally u agreed that it is Ap
anonymous
  • anonymous
so 1/log a , 1/log b and 1/log c are in HP
shubhamsrg
  • shubhamsrg
lol..noo..i didnt.. i meant you can derive loga ,log b and log c are in AP but @sauravshakya 's method..
shubhamsrg
  • shubhamsrg
which ultimately means the ans is HP as you are looking for 1/log a,1/log b and 1/log c
anonymous
  • anonymous
Let: \[\large \log_a(x) = k \implies a^k = x \implies a = x^{\frac{1}{k}}\] Similarly: Let others: \[\large b = x^{\frac{1}{m}}\] \[\large c = x^{\frac{1}{n}}\]
anonymous
  • anonymous
Now given: \[b^2 = ac\]
anonymous
  • anonymous
\[\huge x^{\frac{2}{m}} = x^{\frac{1}{k}} \times x^{\frac{1}{n}}\]
anonymous
  • anonymous
\[\implies \frac{1}{k} + \frac{1}{n} = \frac{2}{m}\] This implies k, m and n are in HP..
anonymous
  • anonymous
How substitute back for k, m and n back..
anonymous
  • anonymous
*now
anonymous
  • anonymous
\[k = \log_a(x), m = \log_b(x), n = \log_c(x)\]
anonymous
  • anonymous
Getting @Yahoo!
anonymous
  • anonymous
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??
anonymous
  • anonymous
This you can do if you want to verify..
anonymous
  • anonymous
i did by this and got answer as it is AP
anonymous
  • anonymous
Let me check then..
anonymous
  • anonymous
can I help?
anonymous
  • anonymous
Yup
anonymous
  • anonymous
|dw:1344689783197:dw| right?
anonymous
  • anonymous
Yup.. Base changing rule
anonymous
  • anonymous
They are in HP @Yahoo!
anonymous
  • anonymous
hw @waterineyes
anonymous
  • anonymous
Let a = 2, b= 4 and c = 8 Let x = 64
anonymous
  • anonymous
ok
anonymous
  • anonymous
|dw:1344689862033:dw|similarly....
anonymous
  • anonymous
right?
anonymous
  • anonymous
@Yahoo!
anonymous
  • anonymous
\[\log_2(64) = 6\] \[\log_4(64) = 3\] \[\log_8(64) = 2\] \[\huge 6, 3, 2 \; \; are \; \; In \; \; HP..\]
anonymous
  • anonymous
if u take x =8
anonymous
  • anonymous
\[\frac{2}{3} = \frac{1}{6} + \frac{1}{2}\]
anonymous
  • anonymous
There is no condition you can take x = 8.. But the fact is that by taking 8 it will not become AP..
anonymous
  • anonymous
log2 (8( = 4 log4(8) = 2 log 8 (8) = 1
anonymous
  • anonymous
\[\huge \log_4(8) \ne 2\]
anonymous
  • anonymous
2^3/2^2 = 2
anonymous
  • anonymous
NO..
anonymous
  • anonymous
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
anonymous
  • anonymous
\[\huge \log_4(8) =\frac{3}{2}\]
anonymous
  • anonymous
Use the base change property that @sauravshakya gave you..
anonymous
  • anonymous
oh..ok.... i really need to study logarithms
anonymous
  • anonymous
@waterineyes @sauravshakya @shubhamsrg thxxxxxx
anonymous
  • anonymous
Welcome dear..

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