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

• Jurrasic

#### 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?

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.

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.

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.