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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 $x$?

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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.

Átila Correia
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latticed-smile
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    You forgot that the logarithm is undefined if its base is 1. – OnTheWay Nov 16 '22 at 02:29
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    Yeah, show $$\lim_{x\to 1}\left( \log_x(2x^{x-2}-1)+4\right)=2.$$ So the second "root" is at $x=1,$ but your function is not actually defined there. – Thomas Andrews Nov 16 '22 at 02:46
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    The function has "holes" in it at $x=1$ and $x=0$ where the function is undefined. You graphing calculator "filled in the holes". This is, in spirit a bit like graphing the function $\frac {x^2-5x+6}{x-3}$ (It looks like the line $y=x-2$) or trying to solve $\frac {x^2-5x+6}{x-3}=1$ (An incorrect solve: $x^2-5x+6=x-3; x^2-6x+9=0;(x-3)^2=0; x=3$ ... but, the function isn't defined at $x=3$ [It's incorrect because by multiplying by $x-3$ I was tacitly assume $x-3\ne 0$ which it of course I can't assume {and by raising $x$ to the $\log_x$ you assume $x$ is a proper log base which it isn't}] ) – fleablood Nov 16 '22 at 04:53

1 Answers1

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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 $x=3$).

fleablood
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