Question

If logab + logcb = 2 (loga b) (logc b), (b not equal to 1) then
(A) a square=bc
(B) b square= ac
(C) c square = ab
(D) 2b = ac

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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

(B)The correct relation is

${b}^{2} = ac$

Step-by-step explanation:

The correct question is "if

$log_{a}b + log_{c}b = 2( log_{ a}b )( log_{c}b)$

Find the relation between a,b &c"

Given logarithmic relation is

$log_{a}b + log_{c}b = 2( log_{ a}b )( log_{c}b)$

We have a basic logarithmic relation,which is mentioned below

$log_{y}(x) = \frac{ logx}{ logy }$

Using this relation for the abive expression,it can be rewritten as follows

$\frac{logb}{loga} + \frac{logb}{logc} = 2 (\frac{logb}{loga}) . (\frac{logb}{logc} ) \\ = > logb( \frac{1}{loga} + \frac{1}{logc} ) = 2 \frac{(logb) {}^{2} }{loga.logc} \\ = > \frac{logc + loga}{loga.logc} = \frac{2logb}{loga.logc} \\ = > logc + loga = 2logb$

We have another two fundamental logarithmic relations which are mentioned below

$logx + logy = log(xy) \\ nlogx = log {x}^{n}$

Substituting the above relations we get the relation as

$log(ac) = log(b {}^{2} ) \\ = > {b}^{2} = ac$

Hence,the relation is found

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