Question

if $y=\left({\log }_{2}3\right)\left({\log }_{3}4\right)....\left({\log }_{31}32\right)$, then

• A. $4<y\le 5$
• B. $y=5$
• C. $4<y<6$
• D. $y=6$

Answer 1

Answer: (A), (B), (C)

$y=5$
$4<y\le 5$
$4<y<6$

Let $y=\left({\log }_{2}3\right)\left({\log }_{3}4\right)\left({\log }_{4}5\right)\dots \left({\log }_{31}32\right)$

Then, $2^y=2^{\left(\right({\log }_{2}3\left)\right({\log }_{3}4\left)\right({\log }_{4}5)\dots ({\log }_{31}32\left)\right)}$

By laws of exponents and the definition of a logarithm,

$2^{\left(\right({\log }_{2}3\left)\right({\log }_{3}4\left)\right({\log }_{4}5)\dots (lo{g}_{31}32\left)\right)}$

$=(2^{\left({\log }_{2}3\right)}{)}^{\left(\right({\log }_{3}4\left)\right({\log }_{4}5)\dots ({\log }_{31}32\left)\right)}$

$=3^{\left(\right({\log }_{3}4\left)\right({\log }_{4}5)\dots ({\log }_{31}32\left)\right)}$

$=(3^{\left({\log }_{3}4\right)}{)}^{\left(\right({\log }_{4}5)\dots ({\log }_{31}32\left)\right)}$

$=4^{\left(\right({\log }_{4}5)\dots ({\log }_{31}32\left)\right)}$ and so on

$={31}^{\left({\log }_{31}32\right)}=32$

$\therefore 2^y=32$

$y=5$