Solve the equation: $\;$ $\log_{0.5x} x^2 - 14 \log_{16x} x^3 + 40 \log_{4x} \sqrt{x} = 0$
Given equation: $\;\;$ $\log_{0.5x} x^2 - 14 \log_{16x} x^3 + 40 \log_{4x} \sqrt{x} = 0$
i.e. $\;$ $2 \log_{0.5x} x - 14 \times 3 \log_{16x} x + 40 \log_{4x} x^{\frac{1}{2}} = 0$
i.e. $\;$ $2 \log_{0.5x} x - 42 \log_{16x} x + 40 \times \dfrac{1}{2} \log_{4x} x = 0$
i.e. $\;$ $\log_{0.5x} x - 21 \log_{16x} x + 10 \log_{4x} x = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_2 x = p$ $\;\;\; \cdots \; (2)$
Now,
$\begin{aligned}
\log_{0.5x} x & = \dfrac{\log_2 x}{\log_2 {0.5x}} \\\\
& = \dfrac{\log_2 x}{\log_2 {0.5} + \log_2 x} \\\\
& = \dfrac{\log_2 x}{\log_2 2^{-1} + \log_2 x} \\\\
& = \dfrac{\log_2 x}{-1 \log_2 2 + \log_2 x} \\\\
& = \dfrac{\log_2 x}{\log_2 x - 1} \\\\
& = \dfrac{p}{p - 1} \;\;\; \cdots \; (3) \;\;\; \left[\text{By equation (2)}\right]
\end{aligned}$
$\begin{aligned}
\log_{16x} x & = \dfrac{\log_2 x}{\log_2 {16x}} \\\\
& = \dfrac{\log_2 x}{\log_2 {16} + \log_2 x} \\\\
& = \dfrac{\log_2 x}{\log_2 2^{4} + \log_2 x} \\\\
& = \dfrac{\log_2 x}{4 \log_2 2 + \log_2 x} \\\\
& = \dfrac{\log_2 x}{\log_2 x + 4} \\\\
& = \dfrac{p}{p + 4} \;\;\; \cdots \; (4) \;\;\; \left[\text{By equation (2)}\right]
\end{aligned}$
$\begin{aligned}
\log_{4x} x & = \dfrac{\log_2 x}{\log_2 {4x}} \\\\
& = \dfrac{\log_2 x}{\log_2 {4} + \log_2 x} \\\\
& = \dfrac{\log_2 x}{\log_2 2^{2} + \log_2 x} \\\\
& = \dfrac{\log_2 x}{2 \log_2 2 + \log_2 x} \\\\
& = \dfrac{\log_2 x}{\log_2 x + 2} \\\\
& = \dfrac{p}{p + 2} \;\;\; \cdots \; (5) \;\;\; \left[\text{By equation (2)}\right]
\end{aligned}$
In view of equations $(3)$, $(4)$ and $(5)$, equation $(1)$ becomes
$\dfrac{p}{p - 1} - \dfrac{21p}{p + 4} + \dfrac{10p}{p + 2} = 0$
i.e. $\;$ $p \left[\dfrac{1}{p - 1} - \dfrac{21}{p + 4} + \dfrac{10}{p + 2}\right] = 0$
$\implies$ $p = 0$ $\;$ or $\;$ $\dfrac{1}{p - 1} - \dfrac{21}{p + 4} + \dfrac{10}{p + 2} = 0$
Case 1: When $\;\;$ $\dfrac{1}{p - 1} - \dfrac{21}{p + 4} + \dfrac{10}{p + 2} = 0$
i.e. $\;$ $\left(p + 4\right) \left(p + 2\right) - 21 \left(p - 1\right) \left(p + 2\right) + 10 \left(p - 1\right) \left(p + 4\right) = 0$
i.e. $\;$ $p^2 + 6p + 8 - 21 p^2 - 21 p + 42 + 10 p^2 + 30 p - 40 = 0$
i.e. $\;$ $10p^2 - 15p - 10 = 0$
i.e. $\;$ $2p^2 - 3p - 2 = 0$
i.e. $\;$ $\left(2p + 1\right) \left(p - 2\right) = 0$
i.e. $\;$ $p = \dfrac{-1}{2}$, $\;$ or $\;$ $p = 2$
Substituting the value of $p$ in equation $(2)$ gives
when $\;\;$ $p = \dfrac{-1}{2}$, $\;$ then $\;$ $\log_2 x = \dfrac{-1}{2}$ $\implies$ $x = 2^{\frac{-1}{2}} = \dfrac{1}{\sqrt{2}}$
when $\;\;$ $p = 2$, $\;$ then $\;$ $\log_2 x = 2$ $\implies$ $x = 2^{2} = 4$
Case 2: When $\;\;$ $p = 0$
$\implies$ $\log_2 x = 0$ $\;\;\;$ [By equation $(2)$]
i.e. $\;$ $x = 2^0 = 1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{1}{\sqrt{2}}, \; 1, \; 4 \right\}$