Loglog a b a-1-log b ab -

The correct option is A 0log { log_ab a+1/log_b ab}= = log { log_ab a+log_ab b } = log (log_ab ab) =log 1=0 ...

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Property 2

loglog a b a+...

Question

$l o g {l o g_{a b} a + \frac{1}{l o g_{b} a b}} =$

0

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l o g a b

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l o g_{a} (a b) = x

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