Trouble factoring with $\ln$

Question

I have the equation

$$0 = 0.5x^\frac {1}{2}(3\ln(x) + 2) $$

How do I interpret the root inside the brackets?

The solutions are $x = 0$ and $x = e^{-\frac{2}{3}}$, but I have absolutely no idea how that last one was found. Could anybody explain it to me?

Answer 1

We have, $$ 0 = 0.5x^\frac {1}{2}(3\ln(x) + 2)$$ Then, $$ 0.5x^\frac {1}{2} \quad \mbox{or}\quad 3\ln(x) + 2=0$$ hence, $$ x=0\quad \mbox{or}\quad \ln(x)=\frac{-2}{3}\Longrightarrow x=e^\frac{-2}{3}$$ And since $\ln(x)$ is only defined on $\Bbb{R}^*_+$ then the only solution is: $$ x=e^\frac{-2}{3}$$

Answer 2

$$0 = 0.5x^{\frac{1}{2}}(3\ln(x)+2)$$

Set either factor equal to $0$. So, you get

$$0.5x^{\frac{1}{2}} = 0 \implies x = 0$$

or

$$3\ln(x)+2 = 0 \implies 3\ln x = -2 \implies \ln x = -\frac{2}{3} \implies x = e^{-\frac{2}{3}}$$

However, since $\ln x$ is defined for all $x > 0$, the first solution is discarded, leaving you with only one solution.

Answer 3

$3 \log x +2=0;$

Since $x >0$ (why?) , we set

$x=e^y$ , $y>0$, real.

Then:

$3 \log (e^y) +2=0;$

$3y +2=0;$

$y=-2/3.$

Since $x=e^y$, we get

$x=e^{-2/3}.$