Question

A denotes the product $x y z$ 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{6 t-3}{4 t-5}=3$
$6 t -3=121-15,6 t =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 $