If $\log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0$, what is the value of $x$?

### Concept & Formula The core concept is converting a logarithmic equation into an exponential equation. If a logarithm equals a number, you can rewrite it as the base raised to that number equals the argument. Formula: If $\log_a(b) = c$, then $b = a^c$. Since this is a nested logarithm, we apply this transformation twice, starting from the outermost base and working our way in. ### Step-by-Step Solution Given the equation: $$ \log_{7} \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 0 $$ Step 1: Remove the outermost logarithm (base 7) by converting it to its exponential form. $$ \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 7^0 $$ Since any non-zero number to the power of 0 is 1, this simplifies to: $$ \log_{5} (\sqrt{x + 5} + \sqrt{x}) = 1 $$ Step 2: Remove the next logarithm (base 5) in the exact same way. $$ \sqrt{x + 5} + \sqrt{x} = 5^1 $$ $$ \sqrt{x + 5} + \sqrt{x} = 5 $$ Step 3: Solve for $x$. The most efficient way is to test the given options to see which one satisfies the simplified equation. Let's test option (c) which is $x = 4$. $$ \sqrt{4 + 5} + \sqrt{4} $$ $$ \sqrt{9} + \sqrt{4} $$ $$ 3 + 2 = 5 $$ This matches our equation perfectly, confirming that $x = 4$ is the correct root. ### Exam Strategy & Shortcut **Option Elimination via Simplification:** Nested log problems appear complex but almost always simplify down to a basic algebraic equation rapidly because the right-hand sides are typically 0 or 1. Once reduced to $\sqrt{x + 5} + \sqrt{x} = 5$, **do not** square both sides mathematically. Option 4 yields perfect squares ($9$ and $4$), making it the most logical choice to test first and saving precious seconds. ### Common Pitfall Students often try to algebraically square the equation $\sqrt{x + 5} + \sqrt{x} = 5$ immediately. Squaring this expression results in $x + 5 + x + 2\sqrt{x^2 + 5x}$, which drastically complicates the problem, creates quadratic equations, and eats up valuable exam time. Always check if plugging in the answers provides a quicker shortcut. ### Final Answer **Therefore, the correct answer is 4.**