Is the solution involving $log$ and linear variables correct? [closed]

Question

$$\ln P + \ln(Dx^4) = - Cx$$

where $P$, $D$ and $C$ are constants. $x$ is the variable of interest which is length parameter (hence $x$is always a non-negative number)

$$\ln P + 4 \ln x + \ln D + Cx = 0$$

Ley $Y = \ln P + \ln D$

$$ 4 \ln x + Cx^2 + Y= 0$$

take derivative w.r.t . x, we get

$$ \frac{4}{x} + 2Cx = 0$$ $$ 4 + 2Cx^2 = 0$$ $$x^2 = \frac{-2}{C}$$

Now that I have $-ve$ sign, I first took square to kill the negative sign, then took the $\sqrt{}$ again. $$x^4 = \frac{4}{C^2} \Rightarrow x = \sqrt{\frac{2}{C}}$$

Answer 1

It may even not be necessary to take the square of $x^2$, then take the 4th root of the result to end up with $x$. For all we know, $C$ could be negative.

In fact, it MUST be that C is negative (by the derivative you calculated) in order to make $x^2$ positive for all $x\geq0$ (since x must be positive due to it being a measure of distance).

Taking the square root of this value gives $x=\pm\sqrt\frac{-2}{C}$ which isn't the same as $x=\sqrt\frac{2}{C}$.

Edit: if the original expression has $-Cx$ on the right hand side, then there is no $x^2$ as Glen points out. In this case, $x=\frac{-4}{C}$ and $C$ still must be negative if $x$ is to be positive.