Question

If $1-{\log }_{x}2+{\log }_{x^2}9-{\log }_{x^3}64<0$, then range of $x$ is $(a,b)$. Find the minimum value of $(a+9b)$

Answer 1

Answer: 25

Clearly $x>0,x\ne 1$
$\therefore 1-{\log }_{x}2+{\log }_{x}3-2{\log }_{x}2<0\text{ or }1-3{\log }_{x}2+{\log }_{x}3<0\text{ or }1-\left({\log }_{x}8-{\log }_{x}3\right)<0$
$1-{\log }_x\left(\frac{8}{3}\right)<0$
If $x>1,{\log }_x\left(\frac{8}{3}\right)>1$. So, $\frac{8}{3}>x\Rightarrow x<\frac{8}{3}$.
Hence $1<x<\frac{8}{3}$
Again if $0<x<1$, then ${\log }_x\left(\frac{8}{3}\right)>1\Rightarrow \frac{8}{3}<x\Rightarrow x>\frac{8}{3}$ (Rejected).
So, $x\in \left(1,\frac{8}{3}\right)\equiv (a,b)$.
$\therefore a=1\text{ and }b=\frac{8}{3}$
Hence $(a+9b)=1+24=25$