if log 5 base 4 = a and log 6 base 5 = b then log 2 base 3 will be ( a) 1/ (2a+1) (b) 1/(2b+1) (c) 2ab+1 (d) 1/(2ab-1)

log45=alog 5log 4=a   ...1      Since logab=log blog alog56=blog 6log 5=b   ...2Multilpying 1 and 2 , we getlog 5log 4log 6log 5=ablog 2×3log 22=ablog 2+log 32 log 2=ablog 2 +log 3=2ab log 22ab log 2-log 2=log 3log 22ab-1=log 3log 2log 3=12ab-1log32=12ab-1Correct answer is D 12ab-1.

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