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I came across this problem: $$ 0 = \ln(c_1 + c_2 \cdot x) + (c_3 + c_4 \cdot x)^{1/2} + c_5 $$

I simplified it down to:

$$ 0 = \ln(c_1 + c_2 \cdot x) + c_3 \cdot x + c_4 $$

Is there a way to solve this analytically?

Sammy Black
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Danny B
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    That simplification is simply not correct (I'm not sure what you are trying to do).... No. I don't think there is any way to solve this analytically at all. You can haave $\ln(c_1 + c_2x) =-(c_3+c_4x)^\frac 12 - c_5 \implies c_1 + c_2 x = e^{-(c_3+c_4x)^\frac 12 - c_5} = \frac 1{e^{\frac 12(c_3+c_4x)}e^{c_5}}$ but I don't think that will do you any good. You simply will not be able to isolate the $x$. You can't simplify an $x$ times a coefficient and an $x$ is an exponent or logarithm – fleablood Jul 20 '21 at 00:05
  • Thanks, yeah I thought I would not be able to isolate the x. My bad, I should've made it clearer that the constants change from one line to the next but still notated the same – Danny B Jul 20 '21 at 00:22

2 Answers2

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You can isolate $x$ only if you make use of special functions (specifically Lambert's product log function $W$):

That may or may not be acceptable to you, depending of what you're trying to do.

For example:

$$ \begin{align} \ln(5+6x)&=2+3x\\ \ln(5+6x)&=\frac{5+6x}{2}-\frac{1}{2}\\ 5+6x&=e^{\frac{5+6x}{2}}\cdot\frac{1}{\sqrt{e}}\\ \frac{5+6x}{2}&=e^{\frac{5+6x}{2}}\cdot\frac{1}{2\sqrt{e}}\\ -\frac{5+6x}{2}\cdot e^{-\frac{5+6x}{2}}&=-\frac{1}{2\sqrt{e}}\\ -\frac{5+6x}{2}&=W\left(-\frac{1}{2\sqrt{e}}\right)\\ x&=-\frac{1}{3}W\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6} \end{align} $$

If solving on real numbers (as opposed to complex), one gets two solutions corresponding to branches $W_0$ and $W_{-1}$:

$$x_1=-\frac{1}{3}W_0\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6}=-\frac{2}{3}$$

and:

$$x_2=-\frac{1}{3}W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6}\approx -0.247856$$

Momo
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Explaining Momo about the Lambert W function. Solution of $$ 0 = \log(c_1+c_2x) + c_3 x + c_4 $$ is $$ x = \frac{1}{c_3}W\left(\frac{c_3}{c_2}\exp \left(\frac{c_1 c_3 - c_4 c_2}{c_2}\right)\right)-\frac{c_1}{c_2} $$ according to Maple.

GEdgar
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