If ${log}_{10} x + {log}_{10} y . \geq 2$ $log_10x+log_10y.>=2$ the minimum value of $x + y =$ $x+y=$ ?

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Answer 1 · 2018-07-16T14:25:28

Given

${log}_{10} x + {log}_{10} y \geq 2$ $log_10x+log_10y>=2$

Here $x > 0 and y > 0$ $x>0andy>0$

$\Rightarrow {log}_{10} (x y) \geq {log}_{10} (10^{2})$ $=>log_10(xy)>=log_10(10^2)$

$\Rightarrow x y \geq 100$ $=>xy>=100$

So minimum value of $x y = 100$ $xy=100$

Now $x + y = {(\sqrt{x} - \sqrt{y})}^{2} + 2 \sqrt{x y}$ $x+y=(sqrtx-sqrty)^2+2sqrt(xy)$

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

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

Hence minimum value of

${(x + y)}_{min} = 10 + 10 = 20$ $(x+y)_"min"=10+10=20$

If log_10x+log_10y.>=2log10x+log10y.≥2log_10x+log_10y.>=2 the minimum value of x+y=x+y=x+y=?