Solving logarithmic equation

**Mathnasium** · April 8th 2008, 07:37 PM

I think I solved this correctly, but could use an extra set of eyes.

$ln(3 \sqrt{x}) = \sqrt{ln x}$ .

Working on the left:

$ln (3 \sqrt{x}) = ln 3 + ln \sqrt{x} = ln 3 + \frac{1}{2}ln{x}$

so now I have:

$ln 3 + \frac{1}{2}ln{x} = \sqrt{ln x}$

I made the substitution $u = \sqrt{ln x}$ , so $u^2 = ln x$ . Performing this substitution gives:

$ln 3 + \frac{1}{2}u^2 = u$ .

Moving everything to one side:

$\frac{1}{2}u^2 - u + ln 3 = 0$ .

A quadratic in u, so using the quadratic formula with a = .5, b = -1, c = ln 3, my discriminant is:

$1 - 2 ln 3$ .

My discriminant is negative, so I have no real solution for u, and thus no real solution for the whole thing.

Look okay? The way the problem was phrased in the text really suggests there's an answer, but...

**o_O** · April 8th 2008, 07:43 PM

Looks good to me

**earboth** · April 8th 2008, 09:16 PM

Originally Posted by Mathnasium

I think I solved this correctly, but could use an extra set of eyes.
...
My discriminant is negative, so I have no real solution for u, and thus no real solution for the whole thing.

Look okay? The way the problem was phrased in the text really suggests there's an answer, but...

All your considerations and calculations are OK.

As an additional information I've attached the graphs of

$f(x) = \ln(3\sqrt{x})$ ...... and ...... $g(x) = \sqrt{\ln(x)}$

The graphs don't intersect that means there doesn't exist a real x so that both terms are equal.

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