# Solving logarithmic equation

• Apr 8th 2008, 07:37 PM
Mathnasium
Solving logarithmic equation
I think I solved this correctly, but could use an extra set of eyes.

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

Working on the left:

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

so now I have:

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

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

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

Moving everything to one side:

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

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

$\displaystyle 1 - 2 ln 3$.

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...
• Apr 8th 2008, 07:43 PM
o_O
Looks good to me (Yes)
• Apr 8th 2008, 09:16 PM
earboth
Quote:

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

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

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