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(B) Given the equation $y = \frac{10^{\log_{10} x}}{x^2}$.Using the logarithmic property $10^{\log_{10} x} = x$,the equation simplifies to $y = \frac{

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$-1$

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(B) Given the equation $y = \frac{10^{\log_{10} x}}{x^2}$.Using the logarithmic property $10^{\log_{10} x} = x$,the equation simplifies to $y = \frac{x}{x^2}$.This further simplifies to $y = \frac{1}{x}$,which implies $x = \frac{1}{y} = y^{-1}$.Given that $x = y^a$,by comparing the exponents,we get $a = -1$.

For $x>0,$ if $y=\frac{10^{\log _{10} x}}{x^{2}}$ and $x=y^{a},$ then $a=$

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