Solve the equation: $\;$ $64^{\frac{1}{x}} - 2^{3 + \frac{3}{x}} + 12 = 0$
Given equation: $\;\;$ $64^{\frac{1}{x}} - 2^{3 + \frac{3}{x}} + 12 = 0$
i.e. $\;$ $\left(8^2\right)^{\frac{1}{x}} - 2^3 \times \left(2^3\right)^{\frac{1}{x}} + 12 = 0$
i.e. $\;$ $\left(8^{\frac{1}{x}}\right)^2 - 8 \times 8^{\frac{1}{x}} + 12 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $8^{\frac{1}{x}} = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 - 8p + 12 = 0$
i.e. $\;$ $\left(p - 6\right) \left(p - 2\right) = 0$
i.e. $\;$ $p = 6$ $\;\;$ or $\;\;$ $p = 2$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = 6$, $\;\;$ $8^{\frac{1}{x}} = 6$
i.e. $\;$ $\dfrac{1}{x} = \log_8 6$
i.e. $\;$ $x = \dfrac{1}{\log_8 6} = \log_6 8 = \log_6 2^3 = 3 \log_6 2$
when $\;$ $p = 2$, $\;$ $8^{\frac{1}{x}} = 2$
i.e. $\;$ $2^{\frac{3}{x}} = 2^1$ $\implies$ $\dfrac{3}{x} = 1$ $\implies$ $x = 3$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \log_6 2, \; 3\right\}$