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Q.
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 $(a+9 b)$
Complex Numbers and Quadratic Equations
Solution:
Clearly $x >0, x \neq 1$
$\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 $(a+9 b)=1+24=25$