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if a,b,c are in G.P then loga (x) , logb (x) , logc (x) are in ?
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shaik0124
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ap
Yahoo!
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Nop ! @shaik0124
shaik0124
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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!
Yahoo!
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But My book says it is H.P
shubhamsrg
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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
sauravshakya
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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
shubhamsrg
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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
sauravshakya
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Thus they are in A.P
shubhamsrg
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exactly!! as @sauravshakya said/did..
waterineyes
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They will be in Harmonic Progression..
Yahoo!
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Yup....they should be in HP ... But hw
Yahoo!
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@waterineyes
shubhamsrg
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ohh,,didnt i explain well @Yahoo! ?
waterineyes
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Let me write..
It will take time.
Yahoo!
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@shubhamsrg finally u agreed that it is Ap
sauravshakya
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so 1/log a , 1/log b and 1/log c are in HP
shubhamsrg
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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..
shubhamsrg
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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
waterineyes
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Let:
\[\large \log_a(x) = k \implies a^k = x \implies a = x^{\frac{1}{k}}\]
Similarly:
Let others:
\[\large b = x^{\frac{1}{m}}\]
\[\large c = x^{\frac{1}{n}}\]
waterineyes
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Now given:
\[b^2 = ac\]
waterineyes
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\[\huge x^{\frac{2}{m}} = x^{\frac{1}{k}} \times x^{\frac{1}{n}}\]
waterineyes
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\[\implies \frac{1}{k} + \frac{1}{n} = \frac{2}{m}\]
This implies k, m and n are in HP..
waterineyes
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How substitute back for k, m and n back..
waterineyes
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*now
waterineyes
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\[k = \log_a(x), m = \log_b(x), n = \log_c(x)\]
waterineyes
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Getting @Yahoo!
Yahoo!
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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??
waterineyes
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This you can do if you want to verify..
Yahoo!
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i did by this and got answer as it is AP
waterineyes
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Let me check then..
sauravshakya
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can I help?
Yahoo!
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Yup
sauravshakya
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|dw:1344689783197:dw|
right?
Yahoo!
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Yup.. Base changing rule
waterineyes
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They are in HP @Yahoo!
Yahoo!
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hw @waterineyes
waterineyes
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Let a = 2, b= 4 and c = 8
Let x = 64
Yahoo!
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ok
sauravshakya
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|dw:1344689862033:dw|similarly....
sauravshakya
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right?
sauravshakya
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@Yahoo!
waterineyes
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\[\log_2(64) = 6\]
\[\log_4(64) = 3\]
\[\log_8(64) = 2\]
\[\huge 6, 3, 2 \; \; are \; \; In \; \; HP..\]
Yahoo!
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if u take x =8
waterineyes
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\[\frac{2}{3} = \frac{1}{6} + \frac{1}{2}\]
waterineyes
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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..
Yahoo!
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log2 (8( = 4
log4(8) = 2
log 8 (8) = 1
waterineyes
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\[\huge \log_4(8) \ne 2\]
Yahoo!
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2^3/2^2 = 2
waterineyes
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NO..
sauravshakya
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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
waterineyes
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\[\huge \log_4(8) =\frac{3}{2}\]
waterineyes
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Use the base change property that @sauravshakya gave you..
Yahoo!
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oh..ok.... i really need to study logarithms
Yahoo!
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@waterineyes @sauravshakya @shubhamsrg thxxxxxx
waterineyes
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Welcome dear..