ln(x)-ln(x+1)=ln(x+3)-ln(x/5)

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Solution for ln(x)-ln(x+1)=ln(x+3)-ln(x/5) equation: 


D( x )

x <= 0

x+3 <= 0

x+1 <= 0

x/5 <= 0

x <= 0

x+3 <= 0

x+3 <= 0

x+3 <= 0 // - 3

x <= -3

x+1 <= 0

x+1 <= 0

x+1 <= 0 // - 1

x <= -1

x/5 <= 0

x/5 <= 0

1/5*x+0 <= 0 // - 0

1/5*x <= 0 // : 1/5

x <= 0/1/5

x <= 0

x in (0:+oo)

ln(x)-ln(x+1) = ln(x+3)-ln(x/5) // - ln(x+3)-ln(x/5)

ln(x)-ln(x+1)-ln(x+3)+ln(x/5) = 0

ln(1*x)-ln(x+1)-ln(x+3)+ln(x/5) = 0

ln((1*x)/(x+1))-ln(x+3)+ln(x/5) = 0

ln(((1*x)/(x+1))/(x+3))+ln(x/5) = 0

ln((x/5)*(((1*x)/(x+1))/(x+3))) = 0

ln((x/5)*((x/(x+1))/(x+3))) = 0

ln((x/5)*((x/(x+1))/(x+3))) = ln(e^0)

(x/5)*((x/(x+1))/(x+3)) = e^0

(x/5)*((x/(x+1))/(x+3))-e^0 = 0

(1/5*x*x)/((x+1)*(x+3))-1 = 0

(1/5*x*x)/((x+1)*(x+3))+(-1*(x+1)*(x+3))/((x+1)*(x+3)) = 0

1/5*x*x-1*(x+1)*(x+3) = 0

-4/5*x^2-4*x-3 = 0

-4/5*x^2-4*x-3 = 0

-1*(4/5*x^2+4*x+3) = 0

4/5*x^2+4*x+3 = 0

DELTA = 4^2-(4/5*3*4)

DELTA = 32/5

DELTA > 0

x = ((32/5)^(1/2)-4)/(4/5*2) or x = (-(32/5)^(1/2)-4)/(4/5*2)

x = 5/8*((32/5)^(1/2)-4) or x = -5/8*((32/5)^(1/2)+4)

-1*(x+5/8*((32/5)^(1/2)+4))*(x-5/8*((32/5)^(1/2)-4)) = 0

(-1*(x+5/8*((32/5)^(1/2)+4))*(x-5/8*((32/5)^(1/2)-4)))/((x+1)*(x+3)) = 0

(-1*(x+5/8*((32/5)^(1/2)+4))*(x-5/8*((32/5)^(1/2)-4)))/((x+1)*(x+3)) = 0 // * (x+1)*(x+3)

-1*(x+5/8*((32/5)^(1/2)+4))*(x-5/8*((32/5)^(1/2)-4)) = 0

( x+5/8*((32/5)^(1/2)+4) )

x+5/8*((32/5)^(1/2)+4) = 0 // - 5/8*((32/5)^(1/2)+4)

x = -(5/8*((32/5)^(1/2)+4))

x = -5/8*((32/5)^(1/2)+4)

( x-5/8*((32/5)^(1/2)-4) )

x-5/8*((32/5)^(1/2)-4) = 0 // + -5/8*((32/5)^(1/2)-4)

x = -(-5/8*((32/5)^(1/2)-4))

x = 5/8*((32/5)^(1/2)-4)

x in { -5/8*((32/5)^(1/2)+4)} 

x in { 5/8*((32/5)^(1/2)-4)} 

x belongs to the empty set

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