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How do you solve #log x+log 20=2#?

1 Answer

Nov 29, 2015

#x=5#

Explanation:

#log(x)+log(20) = log(20x)#

So, if #log(x)+log(20)=2#
then
#color(white)("XXX")log(20x)=2#

#rArrcolor(white)("XX")10^(log(20x)) = 10^2#

#rArrcolor(white)("XX")20x = 100#

#rArrcolor(white)("XX")x=5#

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