```Date: Oct 3, 2012 1:12 PM
Author: Dave L. Renfro
Subject: Re: How to simplify an expression ...

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^(8t-2)]^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 intendedinterpretation, although given the way the spacing was in[ (6 r^8 t) /(-3 r^2) ]^3I still think (r^8)*t is more reasonable than r^(8t).In any event, I also tell students about the sign changemethod. I tell them that when you pass an exponentiatedexpression through a division sign, the algebraic signof the exponent changes (negative to positive or positiveto negative, although almost always you do it in order tochange from negative to positive). Of course, you have topoint out that this is to only be done when the exponentiatedterm 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 processto a rote method rather than something students can seethemselves coming up with on their own. I tend to make moreof the long-and-drawn-out methods I posted about yesterdaywhen a student is first learning this stuff, and put moreemphasis on quickly getting to the desired result withouthaving to think very much when the student is more advancedand the focus is on something else (e.g. calculus studentsrewriting expressions after taking a derivative).Incidentally, you can get students to practice workingintelligentally with exponents by using certain types ofno-calculator 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 10^34You can also do this via (5 x 10^(-5))^(-8), resulting in10^40 / 5^8, at which point you can peal off eight 10's inthe numerator, writing each as 5*2, cancel the eight 5'sin 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 digitsin the base-10 numeral expression for the value of 2^900?90   180   270   450   900   1800Intended 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
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