Question

Number of integers satisfying the inequality $log_2 \sqrt{x} - 2log_{1/4}^2 x + 1 > 0$ is

• A. 1
• B. 2
• C. 3
• D. infinitely many

Answer 1

Answer: (C)

3

Answer 2

The given inequality is $log_2 \sqrt{x} - 2log_{1/4}^2 x + 1 > 0$. The domain requires $x > 0$.

We simplify the logarithmic terms:

1. $log_2 \sqrt{x} = log_2 x^{1/2} = \frac{1}{2} log_2 x$.
2. For $log_{1/4} x$, we use the change of base formula $log_b a = \frac{log_c a}{log_c b}$: $log_{1/4} x = \frac{log_2 x}{log_2 (1/4)} = \frac{log_2 x}{log_2 (2^{-2})} = \frac{log_2 x}{-2} = -\frac{1}{2} log_2 x$. Therefore, $log_{1/4}^2 x = \left(-\frac{1}{2} log_2 x\right)^2 = \frac{1}{4} (log_2 x)^2$.

Substitute these into the inequality: $\frac{1}{2} log_2 x - 2 \left(\frac{1}{4} (log_2 x)^2\right) + 1 > 0$ $\frac{1}{2} log_2 x - \frac{1}{2} (log_2 x)^2 + 1 > 0$

Let $y = log_2 x$. The inequality becomes: $\frac{1}{2} y - \frac{1}{2} y^2 + 1 > 0$

Multiply the inequality by 2 to clear the fractions: $y - y^2 + 2 > 0$

Rearrange into a standard quadratic form and multiply by -1 (reversing the inequality sign): $y^2 - y - 2 < 0$

Factor the quadratic expression: $(y-2)(y+1) < 0$

This inequality holds for values of $y$ between the roots -1 and 2: $-1 < y < 2$

Substitute back $y = log_2 x$: $-1 < log_2 x < 2$

Since the base of the logarithm is $2 > 1$, the exponential function $2^t$ is increasing. Exponentiating with base 2 preserves the inequality direction: $2^{-1} < x < 2^2$ $\frac{1}{2} < x < 4$

We need to find the number of integers satisfying this inequality. The integers in the interval $(\frac{1}{2}, 4)$ are $1, 2,$ and $3$. There are 3 such integers.