CATModern Math > Medium−log2(1/5)-\log _{2}(1 / 5)−log2(1/5)log2(1/5)\log _{2}(1 / 5)log2(1/5)−log2(1/3)-\log _{2}(1 / 3)−log2(1/3)log2(1/3)\log _{2}(1 / 3)log2(1/3)✅ Correct Option: 4Related questions:CAT 2018 Slot 21log2100−1log4100+1log5100−1log10100+1log20100−1log25100+1log50100= ?\dfrac{1}{\log _{2} 100}-\dfrac{1}{\log _{4} 100}+\dfrac{1}{\log _{5} 100}-\dfrac{1}{\log _{10} 100}+\dfrac{1}{\log _{20} 100}-\dfrac{1}{\log _{25} 100}+\dfrac{1}{\log _{50} 100}= \ ?log21001−log41001+log51001−log101001+log201001−log251001+log501001= ?CAT 2021 Slot 3For a real number aaa, if log15a+log32a(log15a)(log32a)=4\frac{\log _{15} a+\log _{32} a}{\left(\log _{15} a\right)\left(\log _{32} a\right)}=4(log15a)(log32a)log15a+log32a=4 then a must lie in the range.CAT 2024 Slot 2If a, b and c are positive real numbers such that a>10≥b≥ca > 10 \ge b \ge ca>10≥b≥c and log8(a+b)log2c+log27(a−b)log3c=23\frac{\log_8(a+b)}{\log_2 c} + \frac{\log_{27}(a - b)}{\log_3 c} = \frac{2}{3}log2clog8(a+b)+log3clog27(a−b)=32, then the greatest possible integer value of a is