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How do I find the solutions of the following equation: $$|x-2|^{10x^2-1}=|x-2|^{3x}\ ?$$

I found that this equation has 5 solutions, 4 positive and 1 negative by looking the graph: enter image description here

Question: How do I compute the values of these roots manually?

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We see that $x=2$ is one solution. Let $x\ne 2$.

Taking $\log$ we get $$(10x^2-1)\log|x-2|=3x\log|x-2|$$

So one solution is $\log |x-2| = 0\implies |x-2| =1 \implies x-2=\pm1 $, so $x=3$ or $x=1$.

Say $\log |x-2| \ne 0$ then $10x^2-1 = 3x$ so $x= {1\over 2}$ and $x=-{1\over 5}$.

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So rearranging gives $$|x-2|^{10x^2-1}-|x-2|^{3x}=0$$ $$|x-2|^{3x}(|x-2|^{10x^2-3x-1}-1)=0$$ So either $x=2$ to achieve zero in the first factor, $|x-2|=1\implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0\implies x=-\frac15 , \frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.

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Hint

Either $$x=2$$or$$|x-2|^{10x^2-3x-1}=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)

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We get easy that $x=2$ is one solution. Now let $x\neq 2$, then it must be $10x^2-1=3x$. Can you finish? Hint: $x=3$ and $x=1$ are also solutions.

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  • $\begingroup$ Yes thank you sir. $\endgroup$ – Namami Shanker Mar 27 at 18:45
  • $\begingroup$ This does not give all of the solutions. $\endgroup$ – Peter Foreman Mar 27 at 18:47
  • $\begingroup$ The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x \ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$. $\endgroup$ – Robert Israel Mar 27 at 18:54

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