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B
12b+1
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C
2ab+1
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D
12ab−1
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Solution
The correct option is D12ab−1 log45=a⇒4a=5log56=b⇒5b=6⇒6=(4a)b=22ab⇒3=22ab2⇒3=22ab−1 Letlog32=x⇒3x=2⋯(1)⇒3x=2(2ab−1)x⋯(2) From (1)and(2)(2ab−1)x=1x=12ab−1