Question

The solution of the equation \( \log\left(\log_4(\sqrt{x+4} + \sqrt{x})\right) = 0 \) is:

• A. \(2\)
• B. \(4\)
• C. \( \frac{9}{4} \)
• D. \(8\)

Hint:

Always verify solution after squaring to avoid extraneous roots.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The equation involves nested logarithms. To solve it, we remove the outer logarithms one by one by converting them into exponential form.
Step 2: Key Formula or Approach:
If \(\log_{b}(y) = a\), then \(y = b^{a}\).
: Detailed Explanation:
Given: \(\log(\log_{4}(\sqrt{x + 4} + \sqrt{x})) = 0\)
Assuming the outer log has a base \(b\) (usually 10 or \(e\), but since the result is 0, the base does not matter as \(b^{0} = 1\)):
\[ \log_{4}(\sqrt{x + 4} + \sqrt{x}) = 1 \]
Now, convert the inner logarithm to exponential form:
\[ \sqrt{x + 4} + \sqrt{x} = 4^{1} = 4 \]
Isolate one square root:
\[ \sqrt{x + 4} = 4 - \sqrt{x} \]
Square both sides:
\[ x + 4 = (4 - \sqrt{x})^{2} \]
\[ x + 4 = 16 + x - 8\sqrt{x} \]
Simplify the equation:
\[ 4 = 16 - 8\sqrt{x} \]
\[ 8\sqrt{x} = 12 \]
\[ \sqrt{x} = \frac{12}{8} = \frac{3}{2} \]
Square again to find \(x\):
\[ x = \left( \frac{3}{2} \right)^{2} = \frac{9}{4} \]
Step 3: Final Answer:
The solution is \(x = 9/4\).