How do you solve #Log(5x)+log(x-1)=2#?

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Answer 1 · 2016-04-19T16:33:31

#x={5}#

#log(5x)+log(x-1)=2#

#"so "log a +log b=log(a*b);#

#"equation can be evaluated as:"#

#log 5x*(x-1)=2#

#if log _a b=c" "then" " b=a^c#

#thus;#

#5x(x-1)=10^2#

#5x(x-1)=100#

#5x^2-5x=100#
#"if both side of equation being divided by 5"#
#x^2-x-20=0#

#(x-5)(x+4)=0#

#if (x-5)=0" ;" rArr x=5#

#if (x+4)=0" ;" rArr x=-4#

#x={5}#