CAT 2022 Slot 2 · Quant · Question 14

The two logarithms have different bases, so convert one to match the other.

Since 4={2}^{2}, {\log}_{4}n=\frac{{\log}_{2}n}{2}.

Write t={\log}_{2}n, so the expression becomes \frac{4-t}{3-\frac{t}{2}}.

Multiplying the denominator out, 3-\frac{t}{2}=\frac{6-t}{2}.

So the fraction is \frac{2\left( 4-t \right)}{6-t}, and the inequality is \frac{4-t}{6-t}<0.

A fraction is negative exactly when numerator and denominator have opposite signs.

For t<4 both 4-t and 6-t are positive, so the fraction is positive — rejected.

For 4<t<6 the numerator is negative and the denominator positive, so the fraction is negative — accepted.

For t>6 both are negative, so the fraction is positive again — rejected.

So 4<t<6, that is 4<{\log}_{2}n<6.

Converting back, {2}^{4}<n<{2}^{6}, so 16<n<64.

The integers strictly between are 17 up to 63, and 63-17+1=47.

Shortcut. Converting to a single base turns the whole thing into \frac{4-t}{6-t}<0, whose solution is simply the gap between the two critical values. Then n runs strictly between {2}^{4} and {2}^{6} — two powers of 2 that are worth reading off directly.