Solve for variable inside and outside of a log 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?
 A: 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:
$$x_2=-\frac{1}{3}W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)-\frac{5}{6}\approx -0.247856$$
A: 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.
