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

[Jurrasic]

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?

 

[eumyang]

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
$$\frac{\log a}{\log b} \ne \log{\left( \frac{a}{b} \right)}$$
... which is what you did on the right side.

 

[HallsofIvy]

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 $y^2- ln(4)y- ln(4)= 0$. 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.

 

[Mark44]

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.