Why is there no real solution: Ln4-Lnx=Ln4/(Lnx)

  • Thread starter Jurrasic
  • Start date
  • #1
92
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and how do you DO THIS?
The question is to find all real roots

steps:
ln(4/x)=ln(4/x)

and then what? Is that right so far?
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
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and how do you DO THIS?
The question is to find all real roots

steps:
ln(4/x)=ln(4/x)

and then what? Is that right so far?
No, this isn't right at all. The left side is fine, but
[tex]\frac{\log a}{\log b} \ne \log{\left( \frac{a}{b} \right)}[/tex]
... which is what you did on the right side.
 
  • #3
HallsofIvy
Science Advisor
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I don't believe that Jurassic meant to say that that was the entire equation- he was just simplifying the left side: ln(x)- ln(4)= ln(x/4)= ln(4)/ln(x) is the entire equation.

But, in fact, he should left it the way it was: ln(x)- ln(4)= ln(4)/ln(x). Now let y= ln(x) and the equation is y- ln(4)= ln(4)/y. Multiply through by y to get the quadratic equation [itex]y^2- ln(4)y- ln(4)= 0[/itex]. Solve that for y= ln(x), using the quadratic formula or by completing the square. Does that quadratic have positive real roots? Remember that the logarithm of any real number must be positive.
 
  • #4
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Remember that the logarithm of any real number must be positive.
I know what HallsOfIvy means, but that's not what he said. (I'm also guilty of misspeaking on occasion.)
The logarithm function is defined only for positive numbers; the result can be positive, negative, or zero. Of course I'm speaking of the usual log function that maps a subset of the reals to the reals.
 

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