Solving this equation: $(x-2)^{x^2-6x+8} >1$, by taking log on base $(x-2)$ both the sides, I get the solution as $x>4$. My work: Let $(x-2)>0$ $$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1 =0\implies x<2, or, x>4$$ But this doesn't appear to be the complete solution for instance $x=5/2$ is also a solution. I would like to know how to solve it completely.

  • $\begingroup$ Can you show your work to get to your solution? $\endgroup$ Jun 4 at 1:24
  • $\begingroup$ Yes, I have included my work now. $\endgroup$ Jun 4 at 1:33
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    $\begingroup$ You'll have to be more careful : If $a^b > 1$ then we can have many possibilities. For example, if $a$ is a negative integer smaller than $-1$ but $b$ is a positive even integer, then the inequality is true. If $a<1$ and $b<0$ then the inequality is true is as well. You'll have to make sure you cover more bases. $\endgroup$ Jun 4 at 1:36
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    $\begingroup$ All even $x\le 0$ is a solution. $\endgroup$
    – Asher2211
    Jun 4 at 1:41
  • $\begingroup$ We can solve $x^2-6x+8>0$ or $(x-3)^2>1$ meaning all solutions $x$ work as long as $x\notin [2,3]$? Then we just need to include any values on this interval that make $x^2-6x+8<0$ or $(x-3)^2<1$ and that would be $x\in (2,4)$...? I'm not sure lol $\endgroup$
    – Mr Pie
    Jun 4 at 3:07

Be careful: when you take the logarithm to the base $x-2$, you need to ensure that $x-2$ is positive. If one of the solutions is $x<2$, you'll be taking a logarithm to a negative base. I would suggest using the natural logarithm instead:



For this quantity to be positive, $(x-2)(x-4)$ and $\ln(x-2)$ must have the same sign. This happens when $2<x<3$ and $x>4$.

PS: Thank you to both Stephen Donovan and Asher2211 for help improving my answer. As I cannot find an expression for all solutions, I will leave my answer as covering the portion with positive $x$.

  • $\begingroup$ This does not help in finding the negative solutions of $ x$ . $\endgroup$
    – Asher2211
    Jun 4 at 1:48
  • $\begingroup$ It should be noted that this gives you all solutions greater than $2,$ for solutions less than $2$ you just need to look for when the exponent is an even integer, which will happen whenever $x$ is even. $\endgroup$ Jun 4 at 1:49
  • $\begingroup$ @StephenDonovan Thank you (and Asher2211) for pointing that out. I will have to fix my answer provided I find a way to come up with other solutions. $\endgroup$
    – Kman3
    Jun 4 at 1:51
  • $\begingroup$ @StephenDonovan Exponent can also be fractions, for example $\frac{10}{3}$ $\endgroup$
    – Asher2211
    Jun 4 at 1:51

Be careful with $f(x)=\log_a x$, apart from $x>0, a>0$, there are two cases when $0 <a<1$ and $a>1$. In the former $f(x)$ decreases and increases in the latter.

So take

Case 1: $0<(x-2)<1 \implies 2<x<3$, then $$(x-2)(x-4)\log_{(x-2)} (x-2) < \log_{(x-2)} 1=0 \implies (x-2)(x-4) <0 \implies 2<x<4.$$ Eventually $2<x<3.$

Case 2: $(x-2)>1 \implies x>3$, then $$(x-2)(x-4)\log_{(x-2)} (x-2) > \log_{(x-2)} 1=0 \implies (x-2)(x-4) >0 \implies x<2, or, x>4.$$ Notice the sign of inequalirt is reversed.So we get $x>4.$

So the total answer is $x\in (2,3) \cup (4,\infty).$


You have covered the case when $x\ge 2$.
It is easy to see that there are no solutions for $1<x<2$.
For $x<1$ all solutions for $x^2-6x+8=\frac{n}{m}$ (where $m$ is odd integer and $n$ is even integer) is a valid solution for the inequality (provided $x<1$) . This is because $a^b$ is real for negative values of $a$ only if $b$ is of the form $\frac{n}{m}$ where $m$ is a odd number and when $n$ is even the $a^b$ attains positive values.


$(x-2)^{x^2-6x+8}\gt 1$


$x^2 -6x + 8=0$ $\implies$ $(x-2)^{x^2-6x+8}=1$

$x\ne4$ and $x\ne2$

At $x=3$


$x\in (2,3)\cup (4,\infty)$

For $x=\frac{5}{2}$



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