- #1

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The question is to find all real roots

steps:

ln(4/x)=ln(4/x)

and then what? Is that right so far?

- Thread starter Jurrasic
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- #1

- 92

- 0

The question is to find all real roots

steps:

ln(4/x)=ln(4/x)

and then what? Is that right so far?

- #2

eumyang

Homework Helper

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No, this isn't right at all. The left side is fine, but

The question is to find all real roots

steps:

ln(4/x)=ln(4/x)

and then what? Is that right so far?

[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

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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

- #4

Mark44

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I know what HallsOfIvy means, but that's not what he said. (I'm also guilty of misspeaking on occasion.)Remember that the logarithm of any real number must be positive.

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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