Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$b^2 = x(1-x) \;\;\implies\;\; x^2 - x + b^2 = 0,$$ which by the quadratic equation, we get $$x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right).$$
My question is: How do I know that neither solution is extraneous? These are the solutions in the back of the textbook, but I am left questioning when these are the solutions.

The alternative solution involves some more clever thinking:
$$\begin{align}2\log_bx + 2\log_b(1-x) &= 4\\\log_bx^2(1-x)^2 &= 4\\ b^4 &= x^2(x^2 - 2x + 1)\\ 0&=x^4 - 2x^3 + x^2 - b^4\\ &=x^2(x - 1)^2 - b^4\\ & = [x(x-1)]^2 - (b^2)^2 \\ &= [x(x-1) + b^2][x(x-1) - b^2],\end{align}$$ which implies that $$x = {1\over2}\left(1 \pm\sqrt{1 + 4b^2}\right) \;\;\;\text{or} \;\;\; {1\over2}\left(1 \pm\sqrt{1 - 4b^2}\right),$$
which is even worse, because now there are $4$ solutions to check.

The most I know is that since $b$ is a logarithmic base, $b>0$ and $b\ne 1$. But what happens when $b$ is something like $2$? Then you end up with a complex number under the root (of the solutions from the easier way), and that doesn't necessarily make sense if I'm trying to solve the logarithmic equation over the reals.

You are correct to transform $2\log_b x + 2\log_b (1-x) = 4$ into $x^2-x+b^2=0$.
These are the restrictions implicated by the original equation:


*

*$b$ is a positive number not equal to $0$ nor $1$

*$x>0 \iff 0<x$

*$1-x>0 \iff 1>x \iff x<1$


I personally felt that it was easiest to wrap my mind around this using a graph with sliders. This shows us that your first equation for the roots is in fact valid.
Going off of that, $x=\dfrac12\left( 1 \pm \sqrt{1-4b^2} \right)$ is defined only when


*

*$1-4b^2\ge 0 \iff 1\ge 4b^2 \iff \frac14 \ge b^2 \iff \frac12\ge|b|\iff |b|\le \frac12$


Now, using all of the aforementioned restrictions (and the graph if you wish), your first solution 
$$\bbox[yellow,5px]{x=\dfrac12\left( 1 \pm \sqrt{1-4b^2} \right)}$$
is correct, and the roots exist only when
$$\bbox[yellow,5px]{0\lt b\le\frac12}$$
(Note that if $b=1/2$, the two roots are the same.)
