Discussion

Explanation:

log₄ log₄ 4^{a - b} = 2log₄ $\left(\sqrt{a}-\sqrt{b}\right)$ + 1

∴ log₄ (a - b) log₄ 4 = 2log₄ $\left(\sqrt{a}-\sqrt{b}\right)$ + 1

∴ log₄ (a - b) = 2log₄ $\left(\sqrt{a}-\sqrt{b}\right)$ + 1

∴ log₄ (a - b) = 2log₄ ${\left(\sqrt{a}-\sqrt{b}\right)}^2$ + 1

∴ log₄ $\left[\frac{a-b}{{\left(\sqrt{a}-\sqrt{b}\right)}^2}\right]$ = 1

∴ $\frac{a-b}{a+b-2\sqrt{ab}}$ = 4

∴ a - b = 4a + 4b - 8$\sqrt{ab}$

∴ 8$\sqrt{ab}$ = 3a + 5b

∴ 8 = 3$\sqrt{\frac{a}{b}}$ + 5$\sqrt{\frac{b}{a}}$

Let $\sqrt{\frac{a}{b}}$ = x

∴ 8 = 3x + $\frac{5}{x}$

∴ 8x = 3x² + 5

∴ 3x² - 8x + 5 = 0

∴ 3x² – 3x – 5x + 5 = 0

∴ 3x(x – 1) – 5(x – 1) = 0

∴(3x – 5)(x – 1) = 0

∴ x = $\frac{5}{3}$ or x = 1

But x ≠ 1 as $\sqrt{a}$ ≠ $\sqrt{b}$

∴ x ≠ 5/3

∴ $\sqrt{\frac{a}{b}}$ = $\frac{5}{3}$

Hence, option (c).

» Your doubt will be displayed only after approval.