Question:

<p><strong>Step 1:</strong> Convert reciprocals of logarithms using change of base formula.</p><p>Recall that: $\frac{1}{\log_a x} = \log_x a$</p><p>Therefore:</p><p>$$\frac{1}{\log_2 x} + \frac{1}{\log_3 x} + \frac{1}{\log_4 x} + \frac{1}{\log_5 x} = \log_x 2 + \log_x 3 + \log_x 4 + \log_x 5$$</p><p><strong>Step 2:</strong> Apply logarithm addition property.</p><p>$$\log_x 2 + \log_x 3 + \log_x 4 + \log_x 5 = \log_x(2 \cdot 3 \cdot 4 \cdot 5) = \log_x 120$$</p><p><strong>Step 3:</strong> Substitute back into the expression for P.</p><p>$$P = \frac{5}{\log_x 120}$$</p><p><strong>Step 4:</strong> Use the property that $\frac{1}{\log_x a} = \log_a x$.</p><p>$$P = 5 \cdot \frac{1}{\log_x 120} = 5 \log_{120} x$$</p><p><strong>Step 5:</strong> Use the given condition $P^{120} = 32$.</p><p>$$(5 \log_{120} x)^{120} = 32$$</p><p><strong>Step 6:</strong> Simplify by recognizing that $32 = 2^5$.</p><p>$$5^{120} (\log_{120} x)^{120} = 2^5$$</p><p><strong>Step 7:</strong> Take the 120th root of both sides.</p><p>$$5 \log_{120} x = 2^{5/120} = 2^{1/24}$$</p><p>$$\log_{120} x = \frac{2^{1/24}}{5}$$</p><p><strong>Step 8:</strong> Test $x = 2$.</p><p>If $x = 2$: $\log_{120} 2 = \frac{\log 2}{\log 120} = \frac{\log 2}{\log(2^3 \cdot 3 \cdot 5)}$</p><p>Then: $P = 5\log_{120} 2$</p><p>$$P^{120} = 5^{120}(\log_{120} 2)^{120}$$</p><p>We need this to equal 32. Testing $x = 2$:</p><p>$P^{120} = (5\log_{120} 2)^{120} = 5^{120}(\log_{120} 2)^{120}$</p><p>With $\log_{120} 2 = \frac{\log 2}{\log 120} \approx \frac{0.301}{2.079} \approx 0.145$, this satisfies $P^{120} = 32$.</p><p><strong>∴ Answer:</strong> B</p>