Logarithms

$\begin{aligned} \large{5^{\sqrt{\frac{\log 7}{\log 5}}}-7^{\sqrt{\frac{\log 5}{\log 7}}}}&=\large{\mathbin{\color{#3D99F6}5}^{\sqrt{\log_{5}7}}-7^{\sqrt{\log_{7}5}}}\\ &=\large{\mathbin{\color{#3D99F6}7}^{\mathbin{\color{#3D99F6}\log_{7}5}\cdot \sqrt{\log_{5}7}}-7^{\sqrt{\log_{7}5}}}\\ &\large{=7^{\sqrt{\log_{7}5}\cdot \sqrt{\mathbin{\color{#D61F06}\log_{7}5\cdot\log_{5}7}}}-7^{\sqrt{\log_{7}5}}}\\ &\large{=7^{\sqrt{\log_{7}5}\cdot \sqrt{\color{#D61F06}1}}-7^{\sqrt{\log_{7}5}}}\phantom{cccccccc}\color{#D61F06}\text{see Note}\\ &\large{=7^{\sqrt{\log_{7}5}}-7^{\sqrt{\log_{7}5}}=\boxed{0}} \end{aligned}$

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Note:

$\log_{a}b\cdot\log_{b}a=\dfrac{\log a}{\log b}\cdot\dfrac{\log b}{\log a}=1$

Let $A=5^{\sqrt{\frac{\log 7}{\log 5}}}$ , $B=7^{\sqrt{\frac{\log 5}{\log 7}}}$ and base of the logarithm be $b$ . Then, we have:

$\begin{aligned} A & = b^{\log A} \\ & = \exp_b \left(\sqrt{\frac{\log 7}{\log 5}} \cdot \log 5 \right) & \small \color{#3D99F6} \text{where }\exp_b (x) = b^x \\ & = \exp_b \left(\sqrt{\log 7\log 5} \right) \end{aligned}$

Similarly, $B = \exp_b \sqrt{\log 5 \log 7} = A$ . $\implies A-B = \boxed{0}$ .