4) \( 72 \cdot\left(49^{\frac{1}{2} \log _{7} 9-\log _{7} 6}+5^{-\log _{\sqrt{5}} 4}\right) \)

To evaluate the expression \( 72 \cdot\left(49^{\frac{1}{2} \log _{7} 9-\log _{7} 6}+5^{-\log _{\sqrt{5}} 4}\right) \), we start simplifying the terms inside the parentheses. 1. Simplifying \( 49^{\frac{1}{2} \log_{7} 9 - \log_{7} 6} \): - We know that \( 49 = 7^2 \). Thus, we can rewrite it as: \[ 49^{\frac{1}{2} \log_{7} 9 - \log_{7} 6} = (7^2)^{\frac{1}{2} \log_{7} 9 - \log_{7} 6} \] - Applying the property of exponents, we get: \[ = 7^{2\left(\frac{1}{2} \log_{7} 9 - \log_{7} 6\right)} = 7^{\log_{7} 9 - 2\log_{7} 6} \] - Using the properties of logarithms, this becomes: \[ = 7^{\log_{7}\left(\frac{9}{6^2}\right)} = \frac{9}{6^2} = \frac{9}{36} = \frac{1}{4} \] 2. Simplifying \( 5^{-\log_{\sqrt{5}} 4} \): - We can change the base using the change of base formula: \[ -\log_{\sqrt{5}} 4 = -\frac{\log_{5} 4}{\log_{5} \sqrt{5}} = -\frac{\log_{5} 4}{\frac{1}{2}} = -2\log_{5} 4 \] - Therefore, \[ 5^{-\log_{\sqrt{5}} 4} = 5^{-2\log_{5} 4} = \left(5^{\log_{5} 4}\right)^{-2} = 4^{-2} = \frac{1}{16} \] 3. Now substituting the simplified terms back into the expression: \[ 72 \cdot \left( \frac{1}{4} + \frac{1}{16} \right) \] - To simplify \( \frac{1}{4} + \frac{1}{16} \), we find a common denominator: \[ \frac{1}{4} = \frac{4}{16} \implies \frac{4}{16} + \frac{1}{16} = \frac{5}{16} \] 4. Thus, our expression now reads: \[ 72 \cdot \frac{5}{16} \] - Calculating this gives us: \[ = \frac{72 \cdot 5}{16} = \frac{360}{16} = 22.5 \] Finally, the value of the expression \( 72 \cdot\left(49^{\frac{1}{2} \log _{7} 9-\log _{7} 6}+5^{-\log _{\sqrt{5}} 4}\right) \) is \(\boxed{22.5}\).