## Homework Statement

Solve for all real x: (x^2-6x+9)^(x^2-4)=1

## The Attempt at a Solution

I raised both sides by 1/x^2-4, so got x^2-6x+9=1, solved it to get x=2,4.
It's a question from a college enterance exam sample paper, so I wondered if it could be that simple. Is my Solution correct?

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Dick
Homework Helper
I'm assuming you didn't mean to put '^' between the two factors. In which case just remember a*b=0 if and only if a=0 or b=0.

It's x^2-6x+9 riased to the power x^2-4

What's $$(x^2-6x+9)^0$$ ?

1. How will that help?

anything to power 0 is 1

so? Ho is it relevant to the ques?

1. How will that help?
Well anything raised to the 0 power is going to equal 1. So, how would you get $$(x^2-6x+9)^{x^2- 4}$$ to be $$(x^2-6x+9)^0$$?

OK I get your point. x=2,-2. But what is wrong with my solution?

OK I get your point. x=2,-2. But what is wrong with my solution?
Well then if you can turn the exponent to 0, then wouldn't the equation also be satisfied when x=-2? (You already have 2 as a solution) Do you see what I'm getting at?

If you don't, graph the equation (x^2-6x+9)^(x^2-4)-1=0 and look at the x-intercepts.

Ok if I take log on both sides then I get (x^2-4)log(x^2-6x+9)=log1=0.
then i get x^2-4=0, for which i get x=2,-2 or x^2-6x+9=1 for which i get x=2,4. so is the answer x=2,-2 or x=2,4?

Ok if I take log on both sides then I get (x^2-4)log(x^2-6x+9)=log1=0.
then i get x^2-4=0, for which i get x=2,-2 or x^2-6x+9=1 for which i get x=2,4. so is the answer x=2,-2 or x=2,4?
Not quite, the answer is going to be x=-2,2,4.

Why not or?
I couldn't graph the func. still haven't learnt enough.

It's going to be x=-2,2,4 because if x=-2 or 2 then the exponent becomes 0, and anything raised to the 0 power will equal 1. x=4 is a solution because:

$$(x^2-6x+9)^{x^2- 4}=1$$
$$((x^2-6x+9)^{x^2- 4})^{1/x^2- 4}=1^{1/x^2- 4}$$
$$x^2-6x+9=1$$
$$(x-2)(x-4)=0$$
$$x=2,4$$

So, 4 is also a solution

Its simple. You get 1 on RHS iff the exponent is 0 OR/AND the base is itself 1.
So, $\ x^2-6x-9=1 \ \mbox{OR/AND} \ x^2-4=0$

Thanks for the help.