Question

A denotes the product $xyz$ where $x,y$ and $z$ satisfy
${\log }_{3}x=\log 5-\log 7$
${\log }_{5}y=\log 7-\log 3$
${\log }_{7}z=\log 3-\log 5$
$B$ denotes the sum of square of solution of the equation
${\log }_2\left({\log }_2{x}^6-3\right)-{\log }_2\left({\log }_2{x}^4-5\right)={\log }_{2}3$
$C$ denotes characterstic of logarithm
${\log }_2\left({\log }_{2}3\right)-{\log }_2\left({\log }_{4}3\right)+{\log }_2\left({\log }_{4}5\right)-{\log }_2\left({\log }_{6}5\right)+{\log }_2\left({\log }_{6}7\right)-{\log }_2\left({\log }_{8}7\right)$
The value of ${\log }_{2}A+{\log }_{2}B+{\log }_{2}C$ is equal to

• A. 5
• B. 6
• C. 7
• D. 4

Answer 1

Answer: (A)

$x=3^{\log 3-\log 7}$
$y=5^{\log 7-\log 3}$
$z=7^{\log 3-\log 5}$
$\therefore x\cdot y\cdot z=1$
$\therefore A=1$
${\log }_2\left(6{\log }_2|x|-3\right)-{\log }_2(4\log |x|-5)={\log }_{2}3$
$\frac{6{\log }_2|x|-3}{4{\log }_2|x|-5}=3\text{ let }{\log }_2|x|=1$
$\therefore \frac{6t-3}{4t-5}=3$
$6t-3=121-15,6t=12$
$\therefore t=2,{\log }_2|x|=2,|x|=4$
$\therefore x=\pm 4$
$B=16+16=32$
${\log }_2\left({\log }_{2}3\right)+{\log }_2\left({\log }_{3}4\right)+{\log }_2\left({\log }_{4}5\right)+{\log }_2\left({\log }_{5}6\right)+{\log }_2\left({\log }_{6}7\right)+{\log }_2\left({\log }_{7}8\right)$
$={\log }_2\left({\log }_{2}8\right)={\log }_{2}3$
$\therefore C=1$