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

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Answer 1 · 2018-07-16T14:25:28

Given

#log_10x+log_10y>=2#

Here #x>0andy>0#

#=>log_10(xy)>=log_10(10^2)#

#=>xy>=100#

So minimum value of #xy=100#

Now #x+y=(sqrtx-sqrty)^2+2sqrt(xy)#

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

So minimum value of y will be such that #y^2=100=>y=10#

Hence minimum value of

#(x+y)_"min"=10+10=20#