# Solving $(x-2)^{x^2-6x+8} >1$

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.

• Can you show your work to get to your solution? Jun 4 at 1:24
• Yes, I have included my work now. Jun 4 at 1:33
• 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. Jun 4 at 1:36
• All even $x\le 0$ is a solution. Jun 4 at 1:41
• 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 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:

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

$$(x-2)(x-4)\ln(x-2)>0$$

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

• This does not help in finding the negative solutions of $x$ . Jun 4 at 1:48
• 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. Jun 4 at 1:49
• @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. Jun 4 at 1:51
• @StephenDonovan Exponent can also be fractions, for example $\frac{10}{3}$ Jun 4 at 1:51

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

So take

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

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 $$1.
For $$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.

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

If

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

$$x\ne4$$ and $$x\ne2$$

At $$x=3$$

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

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

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

$$(0.5)^{-0.75}=1.68179....$$