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Re: How to simplify an expression ...
Posted:
Oct 3, 2012 1:12 PM


Peter Duveen wrote (in part):
http://mathforum.org/kb/message.jspa?messageID=7900080
> I get something a bit different. > > I bring the r^2 up to the top, where it becomes r^2. > I then would tell the student to add the exponents where > the base is the same. I think I interpreted the equation > differently here, making t part of the exponent. > > So, combining the constants, we have [1/2r^(8t2)]^3 > > We cube each factor. That means, in the case of the exponent, > multiplying it by 3. One is left with: > > = 1/16r^(24t  6) > > Did I do something amiss?
I didn't even think of that, and you might have the intended interpretation, although given the way the spacing was in
[ (6 r^8 t) /(3 r^2) ]^3
I still think (r^8)*t is more reasonable than r^(8t).
In any event, I also tell students about the sign change method. I tell them that when you pass an exponentiated expression through a division sign, the algebraic sign of the exponent changes (negative to positive or positive to negative, although almost always you do it in order to change from negative to positive). Of course, you have to point out that this is to only be done when the exponentiated term is factored out of the entire side of the division sign, and after you move it, it's (initially, at least) factored out of the entire side of the other side of the division sign. The method is quick, but it does tend to reduce the process to a rote method rather than something students can see themselves coming up with on their own. I tend to make more of the longanddrawnout methods I posted about yesterday when a student is first learning this stuff, and put more emphasis on quickly getting to the desired result without having to think very much when the student is more advanced and the focus is on something else (e.g. calculus students rewriting expressions after taking a derivative).
Incidentally, you can get students to practice working intelligentally with exponents by using certain types of nocalculator numerical evaluataion problems.
1. (0.00005)^(8) = ??
Intended method: Rewrite as (0.5 x 10^(4))^(8), resulting in (1/2)^(8) x [10^(4)]^(8) = 2^8 x 10^32 = 256 x 10^32 or 2.56 x 10^34
You can also do this via (5 x 10^(5))^(8), resulting in 10^40 / 5^8, at which point you can peal off eight 10's in the numerator, writing each as 5*2, cancel the eight 5's in the numerator with the eight 5's in the denominator, leaving you with eight 2's and 10^32 in the numerator . . .
2. (0.0025)^3 / (0.05)^4 = ??
Intended method: Rewrite as [25 x 10^(4)]^3 / [5 x 10^(2)]^4, which gives [(25)^3 x 10^(12)] / [5^4 x 10^(8)], or (5^6 / 5^4) * (10^8 / 10^12) = . . .
3. Which of the following is closest to the number of digits in the base10 numeral expression for the value of 2^900?
90 180 270 450 900 1800
Intended method: 2^10 = 1024, which is approximately 10^3, so 2^900 = (2^10)^90 is approximately (10^3)^90 = 10^270, so 270 is the closest.
Dave L. Renfro



