${\log _4}18 = . . . .$

${\log _2}7$ એ . . . . થાય.

${\log _3}\,4{\log _4}\,5{\log _5}\,6{\log _6}\,7{\log _7}\,8{\log _8}\,9= . .$ . .

ધારો કે $\quad \sum \limits_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c,$  $a, b, c \in Z$ પુર્ણાકો છે.$e=\sum_{n=0}^{\infty} \frac{1}{n !} $ હોય તો $a^2-b+c$ ની કિમંત મેળવો.

${81^{(1/{{\log }_5}3)}} + {27^{{{\log }_{_9}}36}} + {3^{4/{{\log }_{_7}}9}} = . . . .$

$32\root 5 \of 4 $ to the base $2\sqrt 2 = . . . .$