Logarithm problem 1 by Dhaval Furia

For the equation to have real and distinct roots the discriminant ( $D = b^2 - 4ac$ ) should be greater than 0. Let $\log_{2} {A} = c$ , then:

$\begin{aligned} (-4)^2 -4(1)(-c) &> 0 \\ 4c &< -16 \\ c &> -4 \\ \log_{2} {A} &> -4 \\ A &> \frac{1}{16} \\ \end{aligned}$

Roots of the equation $x^2 - 4x - \text{log}_2 A = 0$ is the abscissa of the points of intersection of the curves $y = x^2 - 4x$ and $y = \text{log}_2 A$ . Now, $y = \text{log}_2 A$ is just the straight line parallel to x-axis. To get more than $1$ points of intersection of these curves, this straight line must intersect the curve $y = x^2 - 4x$ at two points.

$y = x^2 - 4x$ intersects x-axis at points $x = 0$ and $x = 4$ . So the minimum value of the expression $x^2 - 4x$ is attained at $x = \large\frac{0 + 4}{2}$ $= 2$

$\hspace{10pt}\text{log}_2 A > \text{min}(x^2 - 4x)$

$\Rightarrow \text{log}_2 A > 2^2 - 4\cdot 2 = -4$

$\Rightarrow A > 2^{-4}$

$\Rightarrow A > \large\frac{1}{16}$