Question

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

Answer 1

ap

Answer 2

Nop ! @shaik0124

Answer 3

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!

Answer 4

But My book says it is H.P

Answer 5

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

Answer 6

SINCE they are in G.P
b^2 = ac
or, logac/logb = 2
or, loga + log c = 2 logb
now, logb= loga+logc /2

Answer 7

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

Answer 8

Thus they are in A.P

Answer 9

exactly!! as @sauravshakya said/did..

Answer 10

They will be in Harmonic Progression..

Answer 11

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

Answer 12

@waterineyes

Answer 13

ohh,,didnt i explain well @Yahoo! ?

Answer 14

Let me write..
It will take time.

Answer 15

@shubhamsrg finally u agreed that it is Ap

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

Let:

\[\large \log_a(x) = k \implies a^k = x \implies a = x^{\frac{1}{k}}\]

Similarly:

Let others:

\[\large b = x^{\frac{1}{m}}\]

\[\large c = x^{\frac{1}{n}}\]

Answer 20

Now given:

\[b^2 = ac\]

Answer 21

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

Answer 22

\[\implies \frac{1}{k} + \frac{1}{n} = \frac{2}{m}\]

This implies k, m and n are in HP..

Answer 23

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

Answer 24

*now

Answer 25

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

Answer 26

Getting @Yahoo!

Answer 27

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??

Answer 28

This you can do if you want to verify..

Answer 29

i did by this and got answer as it is AP

Answer 30

Let me check then..

Answer 31

can I help?

Answer 32

Yup

Answer 33

right?