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Previous Year Question (PYQs)

For $a>0,\ a\ne1$, the number of values of $x$ satisfying $2\log_x a+\log_{ax} a+3\log_{a^2x} a=0$ is




Solution

Let $\log_x a=t$ Then $\log_{ax}a=\dfrac{t}{1+t}$, $\log_{a^2x}a=\dfrac{t}{2+t}$ Equation becomes $2t+\dfrac{t}{1+t}+3\dfrac{t}{2+t}=0$ Solving gives $t=0,-1,-2$ All give valid $x$ values. Number of solutions $=3$ Answer: $\boxed{3}$

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