Question

Find the value of ${\log }_7{\log }_7$ $\left(7^{\frac{7}{8}}\right)$
A. $1+3{\log }_{7}2$
B. $3{\log }_{2}7$
C. $1-3{\log }_{7}2$
D. ${\log }_{7}3$

• A. 1 + 3 log₇ 2
• B. 3 log₂7
• C. 1 - 3 log₇ 2
• D. 2 log₇ 3

Answer 1

The correct option is C($1-3{\log }_{7}2$)

Given ${\log }_7{\log }_7$ $(7{)}^{\frac{7}{8}}$
Applying the rule of $\log a^m=m\log a$ and also
${\log }_{a}a=1$
Thus, ${\log }_7{\log }_7$ $(7{)}^{\frac{7}{8}}={\log }_7\left(\frac{7}{8}\right){\log }_{7}7$
= ${\log }_7\left(\frac{7}{8}\right)$ (since, ${\log }_{7}7=1$)
Since $\log \left(\frac{a}{b}\right)=\log a-\log b$
= ${\log }_{7}7-{\log }_{7}8$
= 1 - ${\log }_7{2}^3$
= 1 - 3${\log }_{7}2$ (From above rule)