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 $c<y<b,y\ne a$, 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)

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