An algebra problem by Aneesh S.

$\large\log _{ 2 }{ x } =3\quad \Rightarrow \quad x={ 2 }^{ 3 }=8\\ \large\log _{ 8 }{ 64 } =\log _{ 8 }{ { 8 }^{ 2 } } =2\log _{ 8 }{ { 8 } } =\large\color{#20A900}{\boxed { 2 }}$

$\log_x 64 =\log_x 2^6 = 6 \log_x 2 = \frac{6}{\log_2 x} =\frac{6}{3} = \boxed{2}$

Doing it by Logic:

$2$ cubed is $x$ This is $8$ . $64$ expressed in base $8$ is $8^2$ , therefore the value is $2$

$\log _{2}x =3\\x={2}^{3}\\x=8\\ \log _{x}{64}\rightarrow\log _{8}{64} \rightarrow \text{what value (power) to the base 8 gives 64?}\rightarrow 2$

log 2 to the base x=1/3

multiplying 6 on both sides.........

$\log _ a b = c$ , then $a ^ c = b$ .

$\log _ 2 x = 3$ , then $2 ^ 3 = x \to x = 8$ .

$\log _ x 64 = y$ , then $x ^ y = 64 \to 8 ^ y = 64 \to y = \boxed { 2 }$ .

So, the answer is $\boxed2$ .

Log base 2 of x=3 (Log x)/(Log 2)=3 (Log 2)/(Log x) =1/3 6 times [(log 2)/(Log x)]=6 times (1/3) (6 Log 2)/(Log x)=2 (Log (2^6))/(Log x)=2 (Log 64) / (Log x) = 2 Log base x of 64 = 2

Solution $\log_x 64 = \log_x 2^6 = 6 \log_x 2,\text{then:} \frac{6}{ \log_2x} = 6/3 = 2$