For how many integral values of $x$ can a triangle of positive area be formed having side lengths $
\log_{2} x, \log_{4} x, 3$?

$\textbf{(A) } 57\qquad \textbf{(B) } 59\qquad \textbf{(C) } 61\qquad \textbf{(D) } 62\qquad \textbf{(E) } 63$
To determine the number of integral values of \( x \) for which a triangle with side lengths \( \log_{2} x \), \( \log_{4} x \), and \( 3 \) can be formed, we need to apply the triangle inequality. The triangle inequality states that for any triangle with sides \( a \), \( b \), and \( c \):

1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

In our case, the sides of the triangle are \( \log_{2} x \), \( \log_{4} x \), and \( 3 \). We need to ensure that these sides satisfy the triangle inequality.

First, let's express \( \log_{4} x \) in terms of \( \log_{2} x \):
\[
\log_{4} x = \log_{2^2} x = \frac{1}{2} \log_{2} x
\]

Now, we substitute \( a = \log_{2} x \), \( b = \frac{1}{2} \log_{2} x \), and \( c = 3 \) into the triangle inequality conditions:

1. \( \log_{2} x + \frac{1}{2} \log_{2} x > 3 \)
\[
\frac{3}{2} \log_{2} x > 3 \implies \log_{2} x > 2 \implies x > 2^2 \implies x > 4
\]

2. \( \log_{2} x + 3 > \frac{1}{2} \log_{2} x \)
\[
\log_{2} x + 3 > \frac{1}{2} \log_{2} x \implies \frac{1}{2} \log_{2} x + 3 > 0 \implies \log_{2} x > -6
\]
Since \( \log_{2} x > 2 \) from the first inequality, this condition is always satisfied.

3. \( \frac{1}{2} \log_{2} x + 3 > \log_{2} x \)
\[
\frac{1}{2} \log_{2} x + 3 > \log_{2} x \implies 3 > \frac{1}{2} \log_{2} x \implies 6 > \log_{2} x \implies x < 2^6 \implies x < 64
\]

Combining the results from the inequalities, we have:
\[
4 < x < 64
\]

The integral values of \( x \) that satisfy this range are \( 5, 6, 7, \ldots, 63 \). To find the number of these values, we calculate:
\[
63 - 5 + 1 = 59
\]

Thus, there are 59 integral values of \( x \) that satisfy the conditions.

The final answer is \(\boxed{59}\).