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Solve and find the domain of the equation:

 

xlogx(x2)=9

 Feb 12, 2018

Best Answer 

 #2
avatar+26396 
+1

Solve and find the domain of the equation:

xlogx(x2)=9

 

1. Domain:

x2>0x>2

 {xR:x>2} (assuming a function from reals to reals) 

 

2. Solve:

xlogx(x2)=9logx(x2)=ln(x2)ln(x)=ln(x2)ln(x12)=ln(x2)12ln(x)=2ln(x2)ln(x)x2ln(x2)ln(x)=9Formula: ab=eln(ab)=ebln(a)x2ln(x2)ln(x)=e2ln(x2)ln(x)ln(x)=e2ln(x2)

 

e2ln(x2)=9eln((x2)2)=9|eln((x2)2)=(x2)2(x2)2=9| both sidesx2=±3x12=3x1=3+2x1=5x22=3x2=3+2x2=1|no solution, see domain x>2

 

laugh

 Feb 13, 2018
 #1
avatar
+1

Solve for x:
x^(log(x)/(log(sqrt(x)))) = 9

Simplify and substitute y = sqrt(x).
x^(log(x)/(log(sqrt(x)))) = sqrt(x)^4
 = y^4:
y^4 = 9

Taking 4^th roots gives sqrt(3) times the 4^th roots of unity:
y = -sqrt(3) or y = -i sqrt(3) or y = i sqrt(3) or y = sqrt(3)

Substitute back for y = sqrt(x):
sqrt(x) = -sqrt(3) or sqrt(x) = -i sqrt(3) or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or sqrt(x) = -i sqrt(3) or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or x = -3 or sqrt(x) = i sqrt(3) or sqrt(x) = sqrt(3)

Raise both sides to the power of two:
x = 3 or x = -3 or x = -3 or sqrt(x) = sqrt(3)
Raise both sides to the power of two:
x = 3    or x = -3     or x = -3      or x = 3

 

Domain: {x element R : 2 3 + sqrt(5)} (assuming a function from reals to reals)

 Feb 13, 2018
 #2
avatar+26396 
+1
Best Answer

Solve and find the domain of the equation:

xlogx(x2)=9

 

1. Domain:

x2>0x>2

 {xR:x>2} (assuming a function from reals to reals) 

 

2. Solve:

xlogx(x2)=9logx(x2)=ln(x2)ln(x)=ln(x2)ln(x12)=ln(x2)12ln(x)=2ln(x2)ln(x)x2ln(x2)ln(x)=9Formula: ab=eln(ab)=ebln(a)x2ln(x2)ln(x)=e2ln(x2)ln(x)ln(x)=e2ln(x2)

 

e2ln(x2)=9eln((x2)2)=9|eln((x2)2)=(x2)2(x2)2=9| both sidesx2=±3x12=3x1=3+2x1=5x22=3x2=3+2x2=1|no solution, see domain x>2

 

laugh

heureka Feb 13, 2018

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