- anonymous

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

- schrodinger

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- shaik0124

ap

- anonymous

Nop ! @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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## More answers

- anonymous

But My book says it is H.P

- 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

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

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

- anonymous

Thus they are in A.P

- shubhamsrg

exactly!! as @sauravshakya said/did..

- anonymous

They will be in Harmonic Progression..

- anonymous

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

- anonymous

- shubhamsrg

ohh,,didnt i explain well @Yahoo! ?

- anonymous

Let me write..
It will take time.

- anonymous

@shubhamsrg finally u agreed that it is Ap

- anonymous

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

- shubhamsrg

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

- shubhamsrg

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

- 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

Now given:
\[b^2 = ac\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

*now

- anonymous

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

- anonymous

Getting @Yahoo!

- 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

This you can do if you want to verify..

- anonymous

i did by this and got answer as it is AP

- anonymous

Let me check then..

- anonymous

can I help?

- anonymous

Yup

- anonymous

|dw:1344689783197:dw|
right?

- anonymous

Yup.. Base changing rule

- anonymous

They are in HP @Yahoo!

- anonymous

hw @waterineyes

- anonymous

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

- anonymous

ok

- anonymous

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

- anonymous

right?

- anonymous

- anonymous

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

- anonymous

if u take x =8

- anonymous

\[\frac{2}{3} = \frac{1}{6} + \frac{1}{2}\]

- anonymous

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

- anonymous

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

- anonymous

\[\huge \log_4(8) \ne 2\]

- anonymous

2^3/2^2 = 2

- anonymous

NO..

- 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

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

- anonymous

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

- anonymous

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

- anonymous

- anonymous

Welcome dear..

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