If log _{10} x + log _{10} y = 2,then the sma | vedclass

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(C) Given $\log _{10} x + \log _{10} y = 2$.Using the property $\log a + \log b = \log (ab)$,we get $\log _{10} (xy) = 2$.This implies $xy = 10^2 = 10

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$20$

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(C) Given $\log _{10} x + \log _{10} y = 2$.Using the property $\log a + \log b = \log (ab)$,we get $\log _{10} (xy) = 2$.This implies $xy = 10^2 = 100$.By the Arithmetic Mean-Geometric Mean $(AM-GM)$ inequality,for positive real numbers $x$ and $y$,$\frac{x+y}{2} \geq \sqrt{xy}$.Substituting $xy = 100$,we get $\frac{x+y}{2} \geq \sqrt{100}$.$\frac{x+y}{2} \geq 10$.Therefore,$x+y \geq 20$.The smallest possible value of $(x+y)$ is $20$.

If $\log _{10} x + \log _{10} y = 2$,then the smallest possible value of $(x + y)$ is

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