ln(x)-ln(x-2/7x)=ln(5)

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Solution for ln(x)-ln(x-2/7x)=ln(5) equation: 


D( x )

x <= 0

x-((2/7)*x) <= 0

x <= 0

x-((2/7)*x) <= 0

x-((2/7)*x) <= 0

x+(-2/7)*x <= 0

5/7*x+0 <= 0 // - 0

5/7*x <= 0 // : 5/7

x <= 0/5/7

x <= 0

x in (0:+oo)

ln(x)-ln(x-((2/7)*x)) = ln(5) // - ln(5)

ln(x)-ln(x-((2/7)*x))-ln(5) = 0

ln(x)-ln(x+(-2/7)*x)-ln(5) = 0

ln(1*x)-ln(x+(-2/7)*x)-ln(5) = 0

ln((1*x)/(x+(-2/7)*x))-ln(5) = 0

ln((1*x)/(x+(-2/7)*x))+ln(1/5) = 0

ln((1*x)/(x+(-2/7)*x))+ln(1/5) = 0 // - ln(1/5)

ln((1*x)/(x+(-2/7)*x)) = -ln(1/5)

(1*x)/(x+(-2/7)*x) = 1/1/5

(1*x)/(x+(-2/7)*x)-(1/1/5) = 0

x/(5/7*x)-5 = 0

x/(5/7*x)+(-5*5/7*x)/(5/7*x) = 0

x-5*5/7*x = 0

-18/7*x = 0

(-18/7*x)/(5/7*x) = 0

(-18/7*x)/(5/7*x) = 0 // * (5*x)/7

(-18*x)/7 = 0

-18/7*x = 0 // : -18/7

x = 0

x in { 0} 

x belongs to the empty set

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