Room to think …

This post describes what happens when students are allowed to develop their own problem-solving approaches.  Admittedly, this particular situation involves nothing deeper than different forms of symbol manipulation, but I still find it interesting to see how students tackle even simple problems when they aren’t given mandatory methods of solving.

I’m particularly fond of multiple solutions like these.  They encourage students to take approaches that match their thinking and provide spectacular opportunities for classes to compare and contrast different approaches.  Some are elegant, some always work (but require additional algebra), some are just painful–but each offers a different way to think about the problem.  Through my work with CAS, I have always emphasized the importance of recognizing that just because answers look different does not necessarily mean they are different.

Some of our precalculus classes were given the following problem.


Following are 4 different approaches our students used to solve for x, the first three from my classes, and the last from a colleague’s.  (Thanks, Heather!!)

Method I:  One of the most common approaches recognized that logarithm properties can be used to pull down exponents.  These students applied a common logarithm to both sides.

log(5^{x+4}) = log(3^{2x-1})
(x+4)log(5) = (2x-1)log(3)
x log(5)-2x log(3) = -log(3)-4log(5)
x = \frac{-log(3)-4log(5)}{log(5)-2log(3)}

Method II:  A slightly more sophisticated approach applied a logarithm with a base matching one of the given exponentials.  Of course, a base-3 logarithm would work the same way.

log_5(5^{x+4}) = log_5(3^{2x-1})
(x+4) = (2x-1)log_5(3)
x-2x log_5(3) = -log_5(3)-4
x = \frac{-log_5(3)-4}{1-2log_5(3)}

Method III:  Two students decided to employ inverses to change the base of one exponential before equating exponents.

5^{x+4} = (3^{log_3(5)})^{x+4} = 3^{2x-1}
log_3(5)(x+4) = 2x-1
x log_3(5)-2x = -1-4 log_3(5)
x = \frac{-1-4 log_3(5)}{log_3(5)-2}

Method IV:  Yesterday, Heather showed me another approach one of her students used.

5^{x+4} = 3^{2x-1}
5^x 5^4 = 3^{2x} 3^{-1}
5^x (625) = 9^x (\frac{1}{3})
1875 = \frac{9^x}{5^x}
x = log_{\frac{9}{5}}(1875)

This means four very different looking answers were found.  Which were “correct”?  Of course, the question of equality of student responses motivated proof investigations far better than any prompts I could have planned.  It was a good day.


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