Question

If the equation $\frac{\log _{12}\left(\log _8\left(\log _4 x \right)\right)}{\log _5\left(\log _4\left(\log _{ y }\left(\log _2 x \right)\right)\right)}=0$ has a solution for ' $x$ ' when $c < y < b , y \neq a$, where ' $b$ ' is as large as possible and 'c' is as small as possible, then the value of $(a+b+c)$ is equal to

• A. 18
• B. 19
• C. 20
• D. 21

Answer 1

Answer: (B)

$ N ^{ r }=0 ; D ^{ r } \neq 0 \text { and } D ^{ r } \text { is defined; }$
$N ^r=0 \Rightarrow \log _8\left(\log _4 x \right)=1 \Rightarrow \log _4 x =8 \Rightarrow x =2^{16} $
$D^r \neq 0 \Rightarrow \log _4\left(\log _y\left(\log _2 x\right)\right) \neq 1 \Rightarrow \log _y\left(\log _2 x\right) \neq 4 \Rightarrow \log _y(16) \neq 4 $
$16 \neq y^4 \Rightarrow y \neq 2 (y>1 \text {, think } !) \Rightarrow a=2 $
now $D^r$ is defined if and only if $\log _4\left(\log _y\left(\log _2 x\right)\right)>0$
$\Rightarrow \log _{ y }\left(\log _2 x \right)>1 \Rightarrow \log _{ y } 16>1 \Rightarrow 16> y \text {, hence } 1< y <16 $
$\text { comparingwith, } c < y < b \Rightarrow c =1 ; b =16 ; a =2$
$( a + b + c )=19$