Math Carnival Update, and an Algebra Puzzle

Posted on October 18, 2011December 28, 2019 by Denise Gaskins

Oops! I misread my calendar last week. The Math Teachers at Play blog carnival will be this Friday at Maths Insider. That means you still have today and tomorrow to send in your blog post submissions using the handy submission form. See you at the carnival!

In the meantime, let me share with you this monster algebra puzzle from the Well-Trained Mind forum. Simplify:

$[ \left ( {x}^{\frac {3}{2x}} \right )^{\frac{x}{9}} \times \left ( x^{\frac{9}{15}} \right )^{\frac{5}{18}}]^3$

How would you explain this problem to a beginning algebra student who has just learned the exponent rules? Or to his non-mathy mom?

And Don’t Miss…

These other mathy blog carnivals:

7 thoughts on “Math Carnival Update, and an Algebra Puzzle”

Denise says:

October 18, 2011 at 3:56 pm

Oh, bother! I got so interested in that algebra puzzle (and in figuring out how to explain it to the mom) that I let the chicken broth boil away and burnt the bones. No chicken soup for supper tonight — guess it’s time to break out the canned chili…

Reply
Michael Paul Goldenberg says:

October 19, 2011 at 11:31 am

One thing I hope the book or teacher who posed that problem specified is that x is not 0. Otherwise, things will get ugly, as there is an x in the denominator of the fractional exponent on the left.

That aside, are you asking for a solution or for an explanation? The answer, with the non-zero stipulation previously mentioned, is that the whole mess simplifies to x, which no doubt comes as quite a surprise. The first factor simplifies to x^(1/6) and the second to x^(1/6) as well. At this point, you could go one of two ways: 1) multiply inside the square brackets, using the rule for multiplying exponents and get x^(1/3). Then apply the “power to a power” rule and get [x^(1/3)]^3 = x^(3/3) = x^1 = x.

2) Use the “power to a power” rule on everything inside the square brackets, yielding x^(1/2) times x^(1/2), then multiply using the product rule and get x^(1/2 + 1/2) = x^1 = x.

As to why these rules are what they are, I think it’s easiest to illustrate with patterns. See Jim Tanton’s videos on YouTube for a nice job on making some sense of exponent rules. One nice thing about his explanations is that they give you choices by putting things in terms of the notion that “If you like patterns, then it seems to make sense to say that. . . ” followed by whatever he’s showing you. Of course, you could choose NOT to accept what the pattern seems to suggest, but then you’re on your own. . . making your own rules for mathematics and then trying to convince others that your way is the one that makes sense.

Reply
1. Denise says:
  
  October 19, 2011 at 10:15 pm
  
  I haven’t seen the solution, but I imagine they explained the problem with zero. AoPS is usually quite sound. I do know that I myself forgot to mention it — shame on me!
  
  Thank you for the video recommendation. I’m off to look for it now…
  
  Reply
  1. Michael Paul Goldenberg says:
    
    October 20, 2011 at 12:34 pm
    
    Here’s the link: http://www.youtube.com/user/DrJamesTanton#p/u/119/68I5vjZec2Y
    
    Reply
    1. Denise says:
      
      October 21, 2011 at 1:35 pm
      
      Thank you! I think that must be a fairly old video, or else I don’t know how to search YouTube, because I couldn’t find it there. Lots of other cool things (I do love his videos!), but not that one.
      
      Reply
      1. Michael Paul Goldenberg says:
        
        October 21, 2011 at 1:38 pm
        
        Why would you have to search for it? Doesn’t the link take you right there? It does when I click on it. The name of the talk is “Explaining Tricky Exponents”
      2. Denise says:
        
        October 21, 2011 at 1:49 pm
        
        Yes, but when you first posted it, there was no link, so I went searching at his YouTube channel. And then for some unfathomable reason, WordPress.com dumped your link comment in the spam box :(, and I just found it this afternoon.