Discussion

Explanation:

log₇ log₅ $\left(\sqrt{x+5}+\sqrt{x}\right)$ = 0

We know

log_a  b = x ⇒ a^x = b

∴log₅ $\left(\sqrt{x+5}+\sqrt{x}\right)$ = 7⁰
⇒ log₅$\left(\sqrt{x+5}+\sqrt{x}\right)$ = 1
⇒ $\sqrt{x+5}+\sqrt{x}$ = 5¹
⇒ $\sqrt{x+5}=5-\sqrt{x}$

Squaring both sides
⇒ x + 5 = 25 + x − 2 × 5$\sqrt{x}$
⇒ x + 5 = 25 + x + 10$\sqrt{x}$

⇒ 10$\sqrt{x}$ = 20
⇒ $\sqrt{x}$ = 2
⇒ ${\left(\sqrt{x}\right)}^2$ = (2)²
⇒ x = 4

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