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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

$$\mathrm{if}\:\mathrm{log}_{\mathrm{x}} \mathrm{a},\mathrm{log}_{\mathrm{x}\:} \mathrm{b},\mathrm{log}_{\mathrm{x}} \mathrm{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{term}\:\mathrm{of}\:\mathrm{A}.\mathrm{P} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{c}^{\mathrm{2}} =\left(\mathrm{ac}\right)\mathrm{log}_{\mathrm{a}\:} \mathrm{b} \\ $$

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

i apreciate sir! it righty that c^2 =(ab)^(log_a b) !

$$\mathrm{i}\:\mathrm{apreciate}\:\mathrm{sir}!\:\mathrm{it}\:\mathrm{righty}\:\mathrm{that}\:\mathrm{c}^{\mathrm{2}} =\left(\mathrm{ab}\right)^{\mathrm{log}_{\mathrm{a}} \mathrm{b}} ! \\ $$

Commented by 123456 last updated on 07/Nov/16

log_x b−log_x a=log_x c−log_x b(÷log_x a) log_a b−1=log_a c−log_a b 2log_a b=1+log_a c

$$\mathrm{log}_{{x}} {b}−\mathrm{log}_{{x}} {a}=\mathrm{log}_{{x}} {c}−\mathrm{log}_{{x}} {b}\left(\boldsymbol{\div}\mathrm{log}_{{x}} {a}\right) \\ $$$$\mathrm{log}_{{a}} {b}−\mathrm{1}=\mathrm{log}_{{a}} {c}−\mathrm{log}_{{a}} {b} \\ $$$$\mathrm{2log}_{{a}} {b}=\mathrm{1}+\mathrm{log}_{{a}} {c} \\ $$

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

log_(x ) b−log_x a=log_x c−log_(x ) b 2log_(x ) b=log_x a+log_x c log_x c=2log_(x ) b−log_x a =log_x b^2 −log_x a =log_x ((b^2 /a)) c=(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⇒8^2 =(2×8)log_2 4 64=^(?) 16×2 64=^(?) 32 ∴ c^2 ≠(ac)log_a b in general.

$$\mathrm{log}_{\mathrm{x}\:} \mathrm{b}−\mathrm{log}_{\mathrm{x}} \mathrm{a}=\mathrm{log}_{\mathrm{x}} \mathrm{c}−\mathrm{log}_{\mathrm{x}\:} \mathrm{b} \\ $$$$\mathrm{2log}_{\mathrm{x}\:} \mathrm{b}=\mathrm{log}_{\mathrm{x}} \mathrm{a}+\mathrm{log}_{\mathrm{x}} \mathrm{c} \\ $$$$\mathrm{log}_{\mathrm{x}} \mathrm{c}=\mathrm{2log}_{\mathrm{x}\:} \mathrm{b}−\mathrm{log}_{\mathrm{x}} \mathrm{a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}_{\mathrm{x}} \mathrm{b}^{\mathrm{2}} −\mathrm{log}_{\mathrm{x}} \mathrm{a} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{log}_{\mathrm{x}} \left(\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}}\right) \\ $$$$\:\:\:\:\:\:\:\:\mathrm{c}=\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}} \\ $$$$\mathrm{Counter}\:\mathrm{example} \\ $$$$\mathrm{For}\:\mathrm{x}=\mathrm{2},\:\mathrm{a}=\mathrm{2},\mathrm{b}=\mathrm{4},\mathrm{c}=\mathrm{8} \\ $$$$\mathrm{log}_{\mathrm{2}} \mathrm{2}=\mathrm{1}\:,\mathrm{log}_{\mathrm{2}} \mathrm{4}=\mathrm{2}\:\mathrm{and}\:\mathrm{log}_{\mathrm{2}} \mathrm{8}=\mathrm{3}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP} \\ $$$$\mathrm{but}\:\:\mathrm{c}^{\mathrm{2}} =\left(\mathrm{ac}\right)\mathrm{log}_{\mathrm{a}} \mathrm{b}\Rightarrow\mathrm{8}^{\mathrm{2}} =\left(\mathrm{2}×\mathrm{8}\right)\mathrm{log}_{\mathrm{2}} \mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{64}\overset{?} {=}\mathrm{16}×\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{64}\overset{?} {=}\mathrm{32} \\ $$$$\:\therefore\:\:\mathrm{c}^{\mathrm{2}} \neq\left(\mathrm{ac}\right)\mathrm{log}_{\mathrm{a}} \mathrm{b}\:\:\:\:\mathrm{in}\:\mathrm{general}.\:\:\:\: \\ $$

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

but you didn′t show that c^2 =(ac)log_a b!

$$\mathrm{but}\:\mathrm{you}\:\mathrm{didn}'\mathrm{t}\:\mathrm{show}\:\mathrm{that}\:\mathrm{c}^{\mathrm{2}} =\left(\mathrm{ac}\right)\mathrm{log}_{\mathrm{a}} \mathrm{b}! \\ $$

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