if-log-x-a-log-x-b-log-x-c-are-the-term-of-A-P-prove-that-c-2-ac-log-a-b-

Question Number 8893 by j.masanja06@gmail.com last updated on 04/Nov/16

if \log_{x} a, \log_{x} b, \log_{x} c are the term of A . P

prove that c^{2} = (ac) \log_{a} b

Commented by j.masanja06@gmail.com last updated on 04/Nov/16

i apreciate sir! it righty that c^{2} = {(ab)}^{\log_{a} b}!

Commented by 123456 last updated on 07/Nov/16

\log_{x} b - \log_{x} a = \log_{x} c - \log_{x} b (\div \log_{x} a)

\log_{a} b - 1 = \log_{a} c - \log_{a} b

{2 \log}_{a} b = 1 + \log_{a} c

Answered by Rasheed Soomro last updated on 04/Nov/16

\log_{x} b - \log_{x} a = \log_{x} c - \log_{x} b

{2 \log}_{x} b = \log_{x} a + \log_{x} c

\log_{x} c = {2 \log}_{x} b - \log_{x} a

= \log_{x} b^{2} - \log_{x} a

= \log_{x} (\frac{b^{2}}{a})

c = \frac{b^{2}}{a}

Counter example

For x = 2, a = 2, b = 4, c = 8

\log_{2} 2 = 1, \log_{2} 4 = 2 and \log_{2} 8 = 3 are in AP

but c^{2} = (ac) \log_{a} b \Rightarrow 8^{2} = (2 \times 8) \log_{2} 4

64 \overset{?}{=} 16 \times 2

64 \overset{?}{=} 32

∴ c^{2} \neq (ac) \log_{a} b in general .

Commented by j.masanja06@gmail.com last updated on 04/Nov/16

but you {didn}^{'} t show that c^{2} = (ac) \log_{a} b!