Question

If $\log_{9} \log_{5} \left( \sqrt{x+5} + \sqrt{x} \right) = 0$, find the value of $x$.

• A. 1
• B. 0
• C. 2
• D. None of these

Hint:

Always solve nested logarithm problems step-by-step, starting from the outermost log and working inward, ensuring each intermediate result satisfies the domain restrictions.

Answer 1

The Correct Option is A

We are given: $\log_{9} \left[ \log_{5} \left( \sqrt{x+5} + \sqrt{x} \right) \right] = 0$.
Recall that $\log_{a}(y) = 0 \Rightarrow y = 1$.
So from $\log_{9}( \text{something} ) = 0$, we conclude:
$\log_{5} \left( \sqrt{x+5} + \sqrt{x} \right) = 1$.
Now, $\log_{5}(z) = 1 \Rightarrow z = 5$.
Thus: $\sqrt{x+5} + \sqrt{x} = 5$.
Let $\sqrt{x} = t$, then $\sqrt{x+5} = \sqrt{t^2 + 5}$.
Equation becomes: $\sqrt{t^2 + 5} + t = 5$.
Rearrange: $\sqrt{t^2 + 5} = 5 - t$.
Since $\sqrt{t^2 + 5} \ge 0$, we must have $t \le 5$.
Squaring both sides: $t^2 + 5 = (5 - t)^2 = 25 - 10t + t^2$.
Cancel $t^2$ from both sides: $5 = 25 - 10t$.
$-20 = -10t \Rightarrow t = 2$.
Since $t = \sqrt{x}$, we get $\sqrt{x} = 2 \Rightarrow x = 4$.
Wait — this is different from option (A), so let's check carefully.
Recheck: $\sqrt{x+5} + \sqrt{x} = 5$. If $x = 4$, then $\sqrt{9} + \sqrt{4} = 3 + 2 = 5$, correct.
Thus, $\log_{5}(5) = 1$, and $\log_{9}(1) = 0$.
Therefore, $x = 4$, but since this is not in the given options, the correct answer is None of these.