TITLE: How to find the second solution of this $\log_{x}(2x^{x-2} - 1) + 4 = 2x$ logarithmic equation?
QUESTION [4 upvotes]: I have to find the points where:
$$
\log_{x}(2x^{x-2} - 1) + 4 = 2x
$$
It's easy to see that $x=2$ is a solution but, when graphing the function I see that there is another solution around 1. How can I find the value exact value of $x$?

If I try using $x$ as the base to eliminate the logarithm I get the equation:
$$
2x^{x-2} - 1 = x^{2x- 4},
$$
and the solutions to this equation are: $x = 1$ and $x=2$ but $x=1$ cannot be a solution to the first equation.

REPLY [0 votes]: You are correct.  The domain of the function is all positive real numbers other than $1$ and is not defined at $x< 0; x=0;$ or $x=1$.
As such your graphing calculator should have omitted $x=1$ in the graph and graphed the function with "holes".
Your attempt to solve by claiming $\log_x M = K \implies x^K = M$ is making the assumption that $x$ is a legitimate base for a logarithm.  In other words in attempting to solve at all you are immediately beginning by assuming $x>0$ and $x\ne 1$.
So $2x^{x-2} -1 = x^{2x-4}$ does not have such restriction we have slipped in the possibility of extraneous roots.  (Actually it has the restriction that $x \ne 0$ as $0^{neg}$ is undefined but that's another issue).
So when you get $x = 1$ or $x=2$ (actually, how did you get that?  Oh, no matter. It's not pertinent to the point I am trying to make) we must check for extraneous roots.  We must omit any solutions of $x \le 0$ or $x = 1$.
....
This is similar in theory to trying to solve $\frac {x^2-5x +6}{x-3}= 1$ and either doing $\frac {x^2-5x +6}{x-3} = x-2$ (except when $x=3$ where it is undefined) and concluding $x-2 =1$ so $x=3$  (ahem... I said except when $x=3$....) or by doing $x^2-5x+6 = x-3$ so $x^2 -6 + 9 =(x-3)^2 =0$ so $x=3$ (by multiplying both sides by $x-3$ I slipped in the extraneous possibility that $x=3$).