$ \log_{2} 6^{0.5x - 1/4} = 8 $ It is known that
$$ \log_{2} 6^{0.5x - 1/4} = 8 $$
What is $32x$? Put the answer as an integer.

Attempt :
$$ 2^8 = 6^{0.5x - 1/4} \implies 2^{8 \times 64} 6^{16} =6^{32 x} $$
But I cannot seem to write power of 2 in form of power of six. How to find $32x$? Thanks. 

In general, $\log_a b^c = c\log_a b$. Applying that:
$$\log_2 6^{0.5x-1/4} = \left(\dfrac{x}{2}-\dfrac{1}{4} \right)\log_2 6 = 8$$
Divide both sides by $\log_2 6$, and $\dfrac{1}{4}$ to both sides, and multiply both sides by 2:
$$x = 2\left(\dfrac{8}{\log_2 6}+\dfrac{1}{4}\right)$$
$$32x = 64\left( \dfrac{8}{\log_2 6} + \dfrac{1}{4}\right)$$
