How do I get around this problematic $\log$ term? This question came while solving for another question on the site.

Suppose we have a body of mass $m$ with $\vec F_2$ and $\vec F_1$ acting on it as shown above. Let $ \vec F_2= c.(\hat F_2)$ be a constant force, $\vec F_1$ be a variable force, such that, $$P = \vec F_1.\vec v = k$$where $P$ is the instantaneous power corresponding to $\vec F_1$. This power must remain constant. In the situation above, the block is moving towards left, and it's slowing down($\because c>F_1$). Therefore, we can write $$c-F_1 = m\frac{dv}{dt}$$ $$\implies c - F_1 = m\frac{d}{dt}\left(\frac{k}{F_1}\right)$$ After differentiating and rearranging you will get $$\frac{d(F_1)}{dt} = \frac{(F_1)^3}{mk} - \frac{c(F_1)^2}{mk}$$ Now you can integrate by partial fractions(or whatever suits you) to get $$\frac{\ln(F_1 - c)}{c^2}+ \frac{1}{cF_1}-\frac{\ln (F_1)}{c^2} = \frac{t}{mk} + c_1$$ where $c_1$ is the arbitrary constant, which will be removed upon substitution of suitable limits.
Here my idea was that, as $v$ approaches zero, $F_1$ approaches c, thus further lowering the acceleration, such that $F_1.v$ remains equal to $k$, and thus the power due to $F_1$ remains constant, as was asked by the OP in the other question.
But you may notice that '$(F_1 - c)$' will be negative throughout the duration of the motion, and so '$\ln(F_1 - c)$' will not be real.
What is going wrong here? Is my interpretation wrong to begin with? Did my calculation go wrong?
 A: Since everything is in line you can get rid of the vectors and just look at x-axis components.
You will then find the equation of motion as
$$ a= \frac{\rm d}{{\rm d}t} v = \frac{c}{m} -\frac{k}{m v}  $$
The solutions are
$$ \begin{aligned} t & =  \int \frac{1}{a} {\rm d}v = \int_{v_0}^v \frac{m v}{c v-k}{\rm d}v \\ 
  & = \frac{m k}{c^2} \ln \left( \frac{c v-k}{c v_0-k} \right)+\frac{m}{c} (v-v_0)
\end{aligned} \tag{1}$$
$$ \begin{aligned} x & =  \int \frac{v}{a} {\rm d}v = \int_{v_0}^v \frac{m v^2}{c v-k} {\rm d}v \\
 & = \frac{m k^2}{c^3} \ln \left( \frac{c v-k}{c v_0-k} \right) + \frac{m (v^2-v_0^2)}{2 c} + \frac{k m (v-v_0)}{c^2}
\end{aligned} \tag{2} $$
The log functions are valid because their arguments are dimensionless.
The special cases starting from rest, $v_0=0$ then you end up with slightly simplified equations
$$ \begin{aligned} t & = \frac{m k}{c^2} \ln \left( 1- \frac{c v}{k} \right)+\frac{m}{c} v
\end{aligned} \tag{3}$$
$$ \begin{aligned} x & =  \frac{m k^2}{c^3} \ln \left(1- \frac{c v}{k} \right) + \frac{m v^2 }{2 c} + \frac{k m v}{c^2}
\end{aligned} \tag{4} $$
and again the arguments of the log are dimensionless.
Now to invert these equations such that $v(t)$ is derived from (3) and $x( v(t) )$ from (4) it is not possible in this case because there isn't an analytical solution to functions with a variable both inside and outside of a log.
A: Your problem lies in looking only at one term of your equation, ln(F$_1$-c), and in assuming that since this term is not a real number, the equation doesn't make sense. This is caused by two issues.
1 - You did not calculate your constant c$_1$. If you insert t = 0 in your solution, you get that c$_1$ will include for example the term in ln(F$_1$[t=0]-c). This term is also a ln of a negative number. Therefore, in your whole equation, you may very well have complex numbers that have non-zero imaginary part, but these imaginary parts can cancel out so that at the end, you are left with only real numbers. Just like ln(-2) - ln(-2) seems to be indefinite, it makes sens that it is in fact zero. So, the first thing to do would be to get a value for c$_1$ in term of the initial conditions. If you don't know yet what complex numbers are, just concentrate on my ln(-2) example.
2 - The second issue is that you are trying to take the ln of a quantity that has units. What are the units of this kind of term? This doesn't make much sense. Whenever you have a function in an expression that includes physical quantities having units, you want the argument of the function to be dimensionless. Note that in simpler equations, for example x = v t, you can get to dimensionless expressions by rewriting for example as 1 = v t /x. One way to do this with your equations is to combine all the ln terms using the logarithmic addition/subtraction rule. Remember what I said in my first point? By calculating your constant c$_1$, you will get constants in logarithms that will be very similar to your other terms of the solution. You will be able to write dimensionless expressions inside the logarithm, by using the logarithmic addition/subtraction rule, and get to something that looks like what John Alexiou wrote. In doing so, you will get a term of a ln of a ratio of two numbers. If these two numbers are negative, the ratio will be positive, and your ln of a negative number problem will go away.
As he mentioned already, there is no closed form solution to expressions of the type ln(y) + y = x, with y a function of x. In fact, most of 2 variables equations that you can write don't have a solution in which one variable can be expressed as a function of the other. You mention that you are in high school. It just turns out that these solvable equations are the only ones considered in high school, mostly because you cannot do more than solve the other ones numerically, using a computer.
