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Quantitative Aptitude Question on Logarithms

If logalog_a 3030 = AA, logalog_a (53)\bigg(\frac{5}{3}\bigg) = −B-B and log2  alog_2\; a = 13\frac{1}{3}, then log3log_3 aa equals.

A

2A+B−3\frac{2}{A+B}-3

B

A+B−32\frac{A+B-3}{2}

C

2A+B−3\frac{2}{A+B-3}

D

A+B2−3\frac{A+B}{2}-3

Answer

2A+B−3\frac{2}{A+B-3}

Explanation

Solution

The correct answer is (C): 2A+B−3\frac{2}{A+B-3}

loga  5+loga  3+loga  2=Alog_a\;5+log_a\;3+log_a\;2 = A

loga  5+loga  3=A−3log_a\;5+log_a\;3 = A-3

But loga  5−loga  3=−Blog_a\;5-log_a\;3 = -B

Hence 2  loga  3=A+B−32\;log_a\;3 = A+B-3

log3  a=2A+B−3log_3\;a = \frac{2}{A+B-3}