$ \log_{2} 6^{0.5x - 1/4} = 8 $

Question

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.

The solution is not an integer. You can check that graphically by plotting $y=\log_2 6^{\frac{x}{2}-\frac{1}{4}}$. — csch2, Jul 16 '18 at 17:04

score 1 · Accepted Answer · answered Jul 16 '18 at 16:47

1

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:

$$x = 2\left(\dfrac{8}{\log_2 6}+\dfrac{1}{4}\right)$$

$$32x = 64\left( \dfrac{8}{\log_2 6} + \dfrac{1}{4}\right)$$

answered Jul 16 '18 at 16:47

SlipEternal

Thanks. I edited the Q, the result must be in integer. (??) – Redsbefall Jul 16 '18 at 16:49
$$(log_6 8)/32=0.03626745068...$$ I don't know how you would get an integer for x. OOPS, the question has changed since I posted. Never mind. – poetasis Jul 16 '18 at 17:06

1 Answers1