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.
 A: 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.)
A: $\log_b x$ and $\log_b (1-x)$ existing mean $0 < x $ and $0 < 1-x $ so $0 < x < 1$.
When you get $b^2 = x(1-x)$ use similiarity and let $m=|\frac 12 - x|$ so $b^2= (\frac 12 + m)(\frac 12 - m^2) = \frac 14 - m^2 > \frac 14$ so $b < \frac 12$.
Therefore neither solution $x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right)$ is extraneous.
In your second method... I'm not sure why you consider that more clever.  By not dividing by $2$ you are adding extraneous solutions and making it more complicated.  I'd call it a lot less clever.  
But as $0 < x < 1$ you know $x(x-1) + b^2 > 0$ so you can rule out the first pair of solutions.  As as above $b^4 = x^2(1-x)^2$ means $b^2 = x(1-x) < \frac 14$.
A: You have to see that your equation $b^2=x(1-x)$ it has two unknowns actually so there are more than one solution. In fact you have the equation of the circle
$$(x-\frac12)^2+y^2=(\frac12)^2$$
The conclusion is that all the points $(x,b)$ in this semi-circle with $b$ positive are solutions of your problem.
A: You can see from the equation itself that are are, in general, two solutions:  If $x$ is a solution, then so is $1-x$, since $1-(1-x)=x$.  
As for the two spurious solutions $x=(1\pm\sqrt{1+4b^2})/2$ in your second approach, note that for these two, either $x$ or $1-x$ is negative. If the equation were written in the seemingly equivalent form $\log_b(x^2)+\log_b((1-x)^2)=4$, those would indeed be solutions.  But since it's written as $2\log_bx+2\log_b(1-x)=4$, they are not, because you cannot take the log of a negative number.
