Wolfram alpha giving wrong solutions?(link)

[Jarfi]

So yeah, Just using wolfram to help draw a nice function, when this happens:

http://www.wolframalpha.com/input/?i=x^2/ln(x)

It shows a real part under zero, when the function given, x^2/ln(x) has no real solution under zero ! It's domain being:

Dm(f) = ]0;11;infinity[

need explanations fast.

 

[DrClaude]

$$\ln(-x) = \ln(-1 \times x) = \ln(-1) + \ln(x) = i \pi + \ln(x)$$

 

[Jarfi]

$$\ln(-x) = \ln(-1 \times x) = \ln(-1) + \ln(x) = i \pi + \ln(x)$$

This is the imaginary solution.

 

[Student100]

This is the imaginary solution.

I'm confused by the question, but if I understand right what's the of ln(.5), that would give you a negative solution.

If that's not what you mean select real valued plot, and not complex.

 

[Jarfi]

I'm confused by the question, but if I understand right what's the of ln(.5), that would give you a negative solution.

If that's not what you mean select real valued plot, and not complex.

It shows both the real and complex plot, however the real plot is drawn what seems incorrectly, it should stop at zero since x is not defined under zero for the real solution.

 

[Borek]

Then you don't understand what it shows - blue and red are imaginary and real parts of the complex solution, you have to switch to real valued plot (select it from the drop down).

 

[Student100]

It shows both the real and complex plot, however the real plot is drawn what seems incorrectly, it should stop at zero since x is not defined under zero for the real solution.

It's showing you the real and complex parts of the solution. You need to tab down to real valued plot.

 

[Jarfi]

Then you don't understand what it shows - blue and red are imaginary and real parts of the complex solution, you have to switch to real valued plot (select it from the drop down).

Ah, that makes sense. Thanks.