# Solve ln(3*sqrt(x)) = sqrt (ln x)

• Nov 17th 2011, 02:38 PM
fcabanski
Solve ln(3*sqrt(x)) = sqrt (ln x)
I'm hitting a wall on this.

It's:

ln3 + 1/2*ln x = sqrt (ln x)

That's where I hit the wall.
ln 3 + 1/2*u = u^1/2

If I square both sides it gets to be a mess.
• Nov 17th 2011, 02:47 PM
wnvl
Re: Solve ln(3*sqrt(x)) = sqrt (ln x)
$\displaystyle (ln 3 + \frac{u}{2})^2 = u$
$\displaystyle \frac{u^2}{4}+(ln3-1)u+(ln3)^2=0$
...
• Nov 17th 2011, 02:53 PM
fcabanski
Re: Solve ln(3*sqrt(x)) = sqrt (ln x)
And then what? I don't see how to factor that, and tried the quadratic but end up with a complete mess.
• Nov 17th 2011, 03:01 PM
wnvl
Re: Solve ln(3*sqrt(x)) = sqrt (ln x)
Yes, it will be a mess. No alternative.
• Nov 18th 2011, 01:23 AM
sbhatnagar
Re: Solve ln(3*sqrt(x)) = sqrt (ln x)
$\displaystyle \ln{3}+\frac{1}{2}\ln{x}=\sqrt{\ln{x}}$
Quote:

Let $\displaystyle y=\ln{x}$ and $\displaystyle k=\ln{3}$
$\displaystyle \Rightarrow k+\frac{1}{2}y=\sqrt{y}$

$\displaystyle \Rightarrow k^2+\frac{1}{4}y^2+ky=y$.............[Square both the sides]

$\displaystyle \Rightarrow \frac{1}{4}y^2+y(k-1)+k^2=0$

$\displaystyle \Rightarrow y=\frac{(1-k) \pm \sqrt{1-2k}}{\frac{1}{2}}$.............[Quadratic Formula]

Quote:

Re substitute for y and k:
$\displaystyle \Rightarrow \ln{x}=\frac{(1-\ln{3}) \pm \sqrt{1-2\ln{3}}}{\frac{1}{2}}$

$\displaystyle \Rightarrow \ln{x}=2(1-\ln{3}) \pm 2 i \sqrt{2\ln{3}-1}$

$\displaystyle \Rightarrow x=e^{2(1-\ln{3}) + 2 i \sqrt{2\ln{3}-1}}, & e^{2(1-\ln{3}) - 2 i \sqrt{2\ln{3}-1}}$