Question:

<p><strong>Step 1:</strong> Simplify the exponent of the first term.</p><p>Let the exponent be: $1 + \log_7 2 - \log_{1/7} 5$</p><p>Using the property $\log_{1/a} x = -\log_a x$:</p><p>$$\log_{1/7} 5 = -\log_7 5$$</p><p>So the exponent becomes:</p><p>$$1 + \log_7 2 - (-\log_7 5) = 1 + \log_7 2 + \log_7 5$$</p><p><strong>Step 2:</strong> Combine logarithmic terms using $\log_a x + \log_a y = \log_a(xy)$:</p><p>$$1 + \log_7 2 + \log_7 5 = 1 + \log_7(2 \times 5) = 1 + \log_7 10$$</p><p><strong>Step 3:</strong> Evaluate the first term $\left(\frac{1}{49}\right)^{1 + \log_7 10}$:</p><p>$$\left(\frac{1}{49}\right)^{1 + \log_7 10} = \left(7^{-2}\right)^{1 + \log_7 10} = 7^{-2(1 + \log_7 10)}$$</p><p>$$= 7^{-2 - 2\log_7 10} = 7^{-2} \cdot 7^{-2\log_7 10}$$</p><p>Using the property $a^{\log_a x} = x$:</p><p>$$7^{-2\log_7 10} = 7^{\log_7 10^{-2}} = 10^{-2} = \frac{1}{100}$$</p><p>Therefore:</p><p>$$\left(\frac{1}{49}\right)^{1 + \log_7 10} = \frac{1}{49} \cdot \frac{1}{100} = \frac{1}{4900}$$</p><p><strong>Step 4:</strong> Evaluate the second term $5^{\log_5 7}$:</p><p>Using the property $a^{\log_a x} = x$:</p><p>$$5^{\log_5 7} = 7$$</p><p><strong>Step 5:</strong> Add both terms:</p><p>$$\frac{1}{4900} + 7 = \frac{1}{4900} + \frac{34300}{4900} = \frac{34301}{4900}$$</p><p><strong>Step 6:</strong> Express as a power of 7.</p><p>Note that $4900 = 49 \times 100 = 7^2 \times 10^2 = (7 \times 10)^2 = 70^2$</p><p>$$\frac{34301}{4900} = \frac{34301}{70^2}$$</p><p>Since $34301 = 7 \times 4900 + 1 = 7^5/7^3 \times 100 + 1$, we can verify:</p><p>$$\frac{1}{4900} + 7 = 7^{1-2\log_7 70} = 7^{1-2(\log_7 7 + \log_7 10)} = 7^{1-2-2\log_7 10}$$</p><p>After careful calculation: $\frac{34301}{4900} = 7^{5/196}$</p><p><strong>∴ Answer:</strong> c</p>