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

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Answer 1 · 2016-04-19T16:33:31

$x={5}$

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

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

$"equation can be evaluated as:"$

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

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

$thus;$

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

$5x(x-1)=100$

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

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

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

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

$x={5}$

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