Question

Sort the following in the descending order of their values.

• A. 27log35
• B. 4log27
• C. 91+log34
• D. 251+log52

Answer 1

${27}^{{\log }_{3}5}=3^{3\times {\log }_{3}5}$
$=3^{{\log }_3{5}^3}$
$=3^{{\log }_{3}125}$
$=125$

$4^{{\log }_{2}7}=2^{2\times {\log }_{2}7}$
$=2^{{\log }_2{7}^2}$
$=2^{{\log }_{2}49}$
$=49$

$9^{1+{\log }_{3}4}=9^{{\log }_{3}3+{\log }_{3}4}$
$=9^{{\log }_{3}12}=3^{2\times {\log }_{3}12}$
$=3^{{\log }_3{12}^2}$
$=3^{{\log }_{3}144}$
$=144$

${25}^{1+{\log }_{5}2}={25}^{{\log }_{5}5+{\log }_{5}2}$
$={25}^{{\log }_{5}10}=5^{2\times {\log }_{5}10}$
$=5^{{\log }_5{10}^2}$
$=5^{{\log }_{5}100}$
$=100$

Thus, the order is 144>125>100>49

Hence, $9^{1+{\log }_{3}4}>{27}^{{\log }_{3}5}>{25}^{1+{\log }_{5}2}>4^{{\log }_{2}7}$