# If log_10x+log_10y.>=2 the minimum value of x+y=?

Jul 16, 2018

Given

${\log}_{10} x + {\log}_{10} y \ge 2$

Here $x > 0 \mathmr{and} y > 0$

$\implies {\log}_{10} \left(x y\right) \ge {\log}_{10} \left({10}^{2}\right)$

$\implies x y \ge 100$

So minimum value of $x y = 100$

Now $x + y = {\left(\sqrt{x} - \sqrt{y}\right)}^{2} + 2 \sqrt{x y}$

$x + y$ will be minimum when $x = y$

So minimum value of y will be such that ${y}^{2} = 100 \implies y = 10$

Hence minimum value of

${\left(x + y\right)}_{\text{min}} = 10 + 10 = 20$