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If log x base 10 + log y base 10 is greater than or equal to 2 then the smallest possible value of x + y is

Nayak Roshan
44 Points
4 years ago
Hi this  is the problem which you can solve using simple log tricks and AM GM INEQUALITY.
GIVEN logxbase10 + logybase10 is greater than or equal to 2
Logxybase10 is greater than or equal to 2
You must know that lega(base a)is equal to 1 which is the property of log
So you can write 2 as 2×1= 2×log10base10
Now nlogm(any base)=logmn(any base)
So this becomes log102(base 10)
Logxy (bbase 10)is greater than or equal to log100 (base 10)
Cancel log(base 10)on both sides therefore  xy is greater than or equal to 100
By using AM GM INEQUALITY
x+y/2is greater than or equal to (root of (xy))
Therefore if x+y is to be minimum xy must be minimum
Therefore the value of xy is 100
x+y/2 is greater than or equal to (root of (100))
x+y/2 is greater than or equal to 10
WHICH GIVES x+Y IS GREATER THAN OR EQUAL TO 20
THEREFORE MINIMUM VALUE OF x+y is 20.