The number of distinct integer values of n satisfying $\frac{4-\log_2 n}{3-\log_4 n} < 0$, is

For $\frac{4-\log_2 n}{3-\log_4 n} < 0$, the fraction is negative when numerator positive and denominator negative OR numerator negative and denominator positive.

---

First, let us convert $\log_4 n$ to base 2 for easier calculation:

Since $\log_4 n = \frac{\log_2 n}{\log_2 4} = \frac{\log_2 n}{2}$

Our inequality becomes: $\frac{4-\log_2 n}{3-\frac{\log_2 n}{2}} < 0$

Let us substitute $t = \log_2 n$ to simplify:

$\frac{4-t}{3-\frac{t}{2}} < 0$

Multiplying both numerator and denominator by 2:

$\frac{8-2t}{6-t} < 0$

---

Case 1: Numerator positive, denominator negative

$8-2t > 0$ and $6-t < 0$

$t < 4$ and $t > 6$

This is impossible since no number can be both less than 4 and greater than 6.

Case 2: Numerator negative, denominator positive

$8-2t < 0$ and $6-t > 0$

$t > 4$ and $t < 6$

Therefore: $4 < t < 6$

---

Since $t = \log_2 n$:

$4 < \log_2 n < 6$

$2^4 < n < 2^6$

$16 < n < 64$

---

The integers satisfying $16 < n < 64$ are:

$n = 17, 18, 19, 20, ..., 63$

Number of integers = $63 - 17 + 1 = 47$

---

Let us check with $n = 32$ which should satisfy our condition:

$\log_2 32 = 5$ since $2^5 = 32$

$\log_4 32 = \frac{5}{2} = 2.5$

Numerator: $4 - 5 = -1 < 0$

Denominator: $3 - 2.5 = 0.5 > 0$

Fraction: $\frac{-1}{0.5} = -2 < 0$

Answer: 47