ln(x+2)-ln(4x+3)=ln(1/x)

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Solution for ln(x+2)-ln(4x+3)=ln(1/x) equation:


D( x )

4*x+3 <= 0

x = 0

1/x <= 0

x+2 <= 0

4*x+3 <= 0

4*x+3 <= 0

4*x+3 <= 0 // - 3

4*x <= -3 // : 4

x <= -3/4

x = 0

x = 0

1/x <= 0

1/x <= 0

1*x^-1 <= 0

1*x^-1 <= 0 // : 1

x^-1 <= 0/1

x^-1 <= 0

1/(x^1) <= 0

x <> 0

1/(x^1) <= 0 // * x^2

(x^2)/(x^1) <= 0

x^1 <= 0

x <= 0

x in (-oo:0)

x+2 <= 0

x+2 <= 0

x+2 <= 0 // - 2

x <= -2

x in (0:+oo)

ln(x+2)-ln(4*x+3) = ln(1/x) // - ln(1/x)

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

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

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

ln(((1*(x+2))/(4*x+3))/(1/x)) = 0

ln(((x+2)/(4*x+3))/(1/x)) = 0

ln(((x+2)/(4*x+3))/(1/x)) = ln(e^0)

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

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

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

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

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

x-3*x^-1-2 = 0

x-3*x^-1-2 = 0

1*x^1-3*x^-1-2*x^0 = 0

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

x^1*(1*x^2-2*x^1-3*x^0) = 0

x^1

x^2-2*x-3 = 0

x^2-2*x-3 = 0

DELTA = (-2)^2-(-3*1*4)

DELTA = 16

DELTA > 0

x = (16^(1/2)+2)/(1*2) or x = (2-16^(1/2))/(1*2)

x = 3 or x = -1

x in { -1, 3}

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

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

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

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

( x+1 )

x+1 = 0 // - 1

x = -1

( x-3 )

x-3 = 0 // + 3

x = 3

x in { -1}

x = 3

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