Solving Equation: (1/x)^x - x = 0

[EL ALEM]

Homework Statement

(1/x)^x - x = 0

Homework Equations

The Attempt at a Solution

(1/x)^x - x = 0
(1/x)^x = x
xln(1/x) = lnx

Thats all i got up to, I just need a kick in the right direction so I can finish it up.

 

[danago]

Homework Statement

(1/x)^x - x = 0

Homework Equations

The Attempt at a Solution

(1/x)^x - x = 0
(1/x)^x = x
xln(1/x) = lnx

Thats all i got up to, I just need a kick in the right direction so I can finish it up.

How is ln(1/x) related to ln(x)? HINT: Have a look at the properties of logarithms

 

[Outlined]

$$\frac{1}{x^{x}} - x = 0$$

becomes

$$\frac{1}{x^{x}} = x$$

becomes$$1= x^{x+1}$$

becomes

$$\ln 1= \ln x^{x+1}$$

becomes

$$0 = (x+1)\ln x$$

 

[HallsofIvy]

Of course, that has no "elementary" solution. It probably can be solved in terms of Lambert's W function.

 

[danago]

Of course, that has no "elementary" solution. It probably can be solved in terms of Lambert's W function.

What about x=1 or x=-1? Are they not considered elementary solutions; they can be obtained without the use of any non-elementary function?

 

[DCASH88]

(1/x)^x-x=0

(1/x)^x=x

xln(1/x)=lnx

lnx=xln(1/x)

lnx=x(ln1-lnx)

lnx=x(0-lnx)

lnx=-xlnx

x=-1

 

[danago]

(1/x)^x-x=0

(1/x)^x=x

xln(1/x)=lnx

lnx=xln(1/x)

lnx=x(ln1-lnx)

lnx=x(0-lnx)

lnx=-xlnx

x=-1

In the last step you divided both sides by ln(x) which implicitly assumed that ln(x) is not equal to zero i.e. x is not equal to 1. It turns out that x=1 also solves the equation.

 

[berkeman]

(1/x)^x-x=0

(1/x)^x=x

xln(1/x)=lnx

lnx=xln(1/x)

lnx=x(ln1-lnx)

lnx=x(0-lnx)

lnx=-xlnx

x=-1

Please do not do the OP's homework for him. It is against the PF rules (see the link at the top of the page) to provide solutions to homework here. Please confine your help to giving hints, asking questions, finding errors in their work, etc.

 

[DCASH88]

Yes I noticed that just by looking but didn't know how to show. Will you show how to come to that?
Thanks
Daniel

 

[Mentallic]

If you're referring to danago's response, then it's just like solving a quadratic.

When you have $$x^2=x$$ you don't divide through by x to leave x=1, because that implies $$x\neq 0$$, but it is a solution. All you do instead is factorize to leave $$x(x-1)=0$$ which now gives the solutions x=0,1. Do the same for your equality.