Discussing a beginner's exercise (with x in the exponent)

[Dr.Peterson]

How to go about solving this problem? (3x^2-2x)^(6x) = 1

...or get a quadratic:
3x^2 - 2x = 1^(1/(6*x)) : rule: if a^b=c, then a = c^(1/b)
3x^2 - 2x = 1
3x^2 - 2x - 1 = 0

This will not obtain all solutions; don't forget to consider unstated assumptions!

There's a reason the problem asked explicitly for "all real numbers x".

 

[JeffM]

If you take this approach, just be sure to check whether you have lost any possible solutions (or introduced extraneous solutions) by what you have done. For example, taking the log requires some assumptions about the domain, which may or may not be safe. (I've seen other problems like this that have an extra twist.)

What I like about the problem is that it makes you think, rather than just follow familiar paths.

I agree fully, particularly about extraneous solutions and impossible solutions. It is a really good problem.

 

[JeffM]

This will not obtain all solutions; don't forget to consider unstated assumptions!

There's a reason the problem asked explicitly for "all real numbers x".

Maybe I am missing something, but is \(\displaystyle 0^0\) determinate?

As you said, this is a question that requires careful thought.

 

[Dr.Peterson]

Maybe I am missing something, but is \(\displaystyle 0^0\) determinate?

As you said, this is a question that requires careful thought.

I hadn't noticed that little twist; I've been focused on general principles (that definitely apply to other problems of this sort), not on the specific solution of this one.

For some purposes, 0^0 is in fact taken as 1; if I were writing this problem, I probably would have avoided this issue.

And sometimes the careful thought ends up just being confirmation that you didn't miss anything.

 

[JeffM]

For some purposes, 0^0 is in fact taken as 1; if I were writing this problem, I probably would have avoided this issue.

Me too.

 

[mmm4444bot]

… I've been focused on general principles … not on the specific solution …

Oops

For some purposes, 0^0 is in fact taken as 1 …

Not on the Beginning Algebra board.

 

[Dr.Peterson]

Oops


Not on the Beginning Algebra board.

No oops at all, really. I'm not trying to give answers, but to suggest methods.

As for 0^0, I'd say that "indeterminate" is beyond elementary algebra; at this level, we tend to say the x^0 = 1 without worrying about x being 0.

But, again, I'd prefer not to go there.

For the sort of problem I have in mind, consider (2x^2 - 4x + 1)^(x^2 - 8x + 1). I get 5 solutions.

 

[Denis]

Determine all real numbers x such that (3x²-2x)^6x=1
​

Looks more like a trick question to me.
That "6x" could be replaced by anything you wish:
If u^(anything) = 1, then u HAS TO equal 1.
So 3x - 2x = 1. Over and out...

My opinion of course, and I respect yours,
but if it's not the same as mine, then it's wrong

Seriously: let's not forget that the OP is "learning"
about powers...

 

[JeffM]

As for 0^0, I'd say that "indeterminate" is beyond elementary algebra; at this level, we tend to say the x^0 = 1 without worrying about x being 0.

But, again, I'd prefer not to go there.

When I teach high school students about the laws of exponents, I always stipulate a positive base. Therefore, I do not talk about 0^0 being indeterminate because I do not talk about 0^q at all.

The advantage in talking exclusively about a positive base a is that it greatly simplifies defining fractional exponents and motivating the definition of \(\displaystyle a^0 \equiv 1.\)

Obviously, \(\displaystyle a^0 = \dfrac{a^n}{a^n} = 1\)

is problematic if a = 0.

I doubt that a question about

\(\displaystyle \{f(x) + g(x)\}^{h(x)} = y\)

should be posted in beginning algebra, at least not if we are considering beginning algebra to mean first-year algebra in US high schools. So I am uncertain whether indeterminacy is relevant or not.

 

[JeffM]

Looks more like a trick question to me.
That "6x" could be replaced by anything you wish:
If u^(anything) = 1, then u HAS TO equal 1.
So 3x - 2x = 1. Over and out...

My opinion of course, and I respect yours,
but if it's not the same as mine, then it's wrong

Seriously: let's not forget that the OP is "learning"
about powers...

Mon ami, \(\displaystyle u > 0 \implies u^0 = 1 \text { even if } u \ne 1.\)

I am doubtful that this is a problem about teaching the basics of powers. It looks more like a problem about working with exponential and logarithmic equations.

 

[mmm4444bot]

… As for 0^0, I'd say that "indeterminate" is beyond elementary algebra …

But, again, I'd prefer not to go there.

I can understand why. By the way:

0^0 = Undefined

This is one of the exponent properties taught in elementary algebra.

Additionally, as a beginner's method, raising each side of an equation by the reciprocal of some exponent is also taught.

If x = 0 were a student's candidate for the solution set, I would hope they've already learned the exponent properties and been trained to check their answers.

 

[mmm4444bot]

I don't understand how people are getting their data. We don't know from where this exercise comes or what classrooms are doing, in general.

I see good methods posted, in this thread. Given a choice, I like Denis' approach (#6). If the expectation is to use logarithms, JeffM's approach is good (#3).

 

[Steven G]

I'm new here, so hello everybody!

Anyways, I've been struggling a lot with questions where there are variables in the exponent.

How would I solve this problem? (And what would you call a problem like this, so I could try to find more for practice.)

Determine all real numbers x such that (3x²-2x)^6x=1


​Thanks!

This is one of the best problems I have seen in a while. Solving it is a matter of careful thinking. In my opinion using logs can complicate things as it brings in additional things to think about.

 

[Steven G]

Looks more like a trick question to me.
That "6x" could be replaced by anything you wish:
If u^(anything) = 1, then u HAS TO equal 1.
So 3x - 2x = 1. Over and out...

I disagree with your over and out. How about u^0 = 1. Why are you not considering that possibility?

 

[lookagain]

One would already establish that x cannot equal 0, so the exponent 6x cannot equal 0. But, as a part of solving this, you must check to see if the base \(\displaystyle \ (3x^2 - 2x) \ \)
could equal -1, with the exponent 6x equaling some even integer, negative or positive, at the same time, so that the result would equal 1. At least you would check this case.