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Solve the equation :

In(5-x) = In5 - Inx,

giving your answers correct to 3 significant figures.

Sagot :

[tex]\ln(5-x) = \ln 5 - \ln x\\\\\implies \ln(5-x) + \ln x = \ln 5\\\\\implies \ln[(5-x)x] =\ln 5\\\\\implies \ln(5x -x^2) = \ln 5\\\\\implies 5x -x^2 = 5\\\\\implies x^2 -5x +5 =0\\\\\implies x = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 5}}{2(1)}\\\\\implies x = \dfrac{5 \pm \sqrt{5}}{2}\\\\\text{Hence} ~x = \dfrac 12\left(5 + \sqrt 5\right)=3.62~~ \text{and}~~ x =\dfrac 12 \left(5-\sqrt 5\right) =1.38[/tex]

Answer:

x1 = 3.618, x2 = 1.382

Step-by-step explanation:

ln(5-x) = ln5 - lnx

ln(5-x) + ln(x) = ln5

ln(5x-x^2) = ln5

5x-x^2=5

0=x^2-5x+5

x = -(-5)±√((-5)²-4(1)(5)) / 2(1)

x = 5±√(25-20) / 2

x = 5±√5 / 2

x1 = 3.618, x2 = 1.382