5^{sqrt{log _{5} 7}} - 7^{sqrt{log _{7} 5}} | vedclass

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(C) Let $x = 5^{\sqrt{\log _{5} 7}}$ and $y = 7^{\sqrt{\log _{7} 5}}$.We know the property $a^{\log_{a} b} = b$.Consider the term $y = 7^{\sqrt{\log _

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(C) Let $x = 5^{\sqrt{\log _{5} 7}}$ and $y = 7^{\sqrt{\log _{7} 5}}$.We know the property $a^{\log_{a} b} = b$.Consider the term $y = 7^{\sqrt{\log _{7} 5}}$.Since $\log_{7} 5 = \frac{1}{\log_{5} 7}$,we can write:$y = 7^{\sqrt{\frac{1}{\log_{5} 7}}} = 7^{\frac{1}{\sqrt{\log_{5} 7}}}$.Using the property $a^{\log_{a} b} = b$,we know $7 = 5^{\log_{5} 7}$.Substituting this into $y$:$y = (5^{\log_{5} 7})^{\frac{1}{\sqrt{\log_{5} 7}}} = 5^{\log_{5} 7 \cdot \frac{1}{\sqrt{\log_{5} 7}}} = 5^{\sqrt{\log_{5} 7}}$.Thus,$x = y$,which implies $x - y = 0$.

$5^{\sqrt{\log _{5} 7}} - 7^{\sqrt{\log _{7} 5}}$

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