# Monthly Archives: December 2012

I received a lovely math mug as a gift from one of my students this year.  Thank you, E.

I put it in the dishwasher before using it.  Taking it out, I found a funny note from the manufacturer on the bottom of the mug.

My question:  What did someone do to inspire that indelible message on the bottom of coffee mug?

Happy Holidays, every one.

## Mobius Strip for Chanukah

A student of mine, S, works with clay in the summer.  This past summer, she made a Mobius Strip Menorah in the spirt of Vi Hart.

I think this is also a great addition to the mathematical art that Patrick Honner is always sharing in the Twitterverse.

Happy Holidays everyone!

## Exponentials and Transformations

Here’s an old and (maybe) a new way to think about equations of exponential functions.  I suspect you’ve seen the first approach.  If you understand what exponentials functions are, my second approach using transformations is much faster and involves no algebra!

Members of the exponential function family can be written in the form $y=a\cdot b^x$ for real values of a and postive real values of b.  Because there are only two parameters, only two points are required to write an equation of any exponential.

EXAMPLE 1: Find an exponential function through the points (2,5) and (4,20).

METHOD 1:  Plug the points into the generic exponential equation to get a 2×2 system of equations.  It isn’t necessary, but to simplify the next algebra step, I always write the equation with the larger exponent on top.

$\left\{\begin{matrix} 20=a\cdot b^4 \\ 5=a\cdot b^2 \end{matrix}\right.$

If the algebra isn’t the point of the lesson, this system could be solved with a CAS.  Users would need to remember that $b>0$ to limit the CAS solutions to just one possibility.

If you want to see algebra, you could use substitution, but I recommend division.  Students’ prior experience with systems typically involved only linear functions for which they added or subtracted the equations to eliminate variables.  For exponentials, the unknown parameters are multiplied, so division is a better operational choice.  Using the system above, I get $\displaystyle \frac{20}{5}=\frac{a\cdot b^4}{a\cdot b^2}$.  The fractions must be equivalent because their numerators are equal and their denominators are equal.

Simplifying gives $4=b^2\rightarrow b=+2$ (because $b>0$ for exponential functions) and $a=\frac{5}{4}$.

This approach is nice because the a term will always cancel from the first division step, leaving a straightforward constant exponent to undo, a pretty easy step.

METHOD 2:  Think about what an exponential function is and does.  Then use transformations.

Remember that linear functions ($y=m\cdot x+b$) “start” with a y-value of b (when $x=0$) and add m to y every time you add 1 to x.  The only difference between linear and exponential functions is that exponentials multiply while linears add.  Therefore, exponential functions ($y=a\cdot b^x$) “start” with a y-value of a when $x=0$ and multiply by b every time 1 is added to x.

What makes the given points a bit annoying is that neither is a y-intercept.  No problem.  If you don’t like the way a problem is phrased, CHANGE IT!    (Just remember to change the answer back to the original conditions!)

If you slide the given points left 2 units, you get (0,5) and (2,20).  It would also be nice if the points were 1 x-unit apart, so halving the x-values gives (0,5) and (1,20).  Because the y-intercept is now 5, and the next point multiplies that by 4, an initial equation for the exponential is $y = 5\cdot 4^x$ . To change this back to the original points, undo the transformations at the start of this paragraph:  stretch horizontally by 2 and then move right 2.  This gives $y = 5\cdot 4^\frac{x-2}{2}$.

This is algebraically equivalent to the $y=\frac{5}{4}\cdot 2^x$ found early.  Obviously, my students prove this.

One student asked why we couldn’t make the (4,20) point the y-intercept.  Of course we can!  To move more quickly through the set up, starting at (4,20) and moving to (2,5) means my initial value is 20 and I multiply by $\frac{1}{4}$ if the x-values move left 2 from an initial x-value of 4.  This gives $y = 20\cdot\left( \frac{1}{4} \right) ^\frac{x+4}{-2}$.  Of course, this 3rd equation is algebraically equivalent to the first two.

Here’s one more example to illustrate the speed of the transformations approach, even when the points aren’t convenient.

EXAMPLE 2: Find an exponential function through (-3,7) and (12,13).

Starting at (-3,7) and moving to (12,13) means my initial value is 7, and I multiply by $\frac{13}{7}$ if the x-values move right 15 from an initial x-value of -3.  This gives $y = 7\cdot\left( \frac{13}{7} \right) ^\frac{x-3}{15}$.

Equivalently, starting at (12,13) and moving to (-3,7) means my initial value is 13, and I multiply by $\frac{7}{13}$ if the x-values move left 15 from an initial x-value of 12.  This gives $y = 13\cdot\left( \frac{7}{13} \right) ^\frac{x+3}{-15}$.

If you get transformations, exponential equations require almost no algebraic work, no matter how “ugly” the coordinates.  I hope this helps give a different perspective on exponential function equations and helps enhance the importance of the critical math concept of equivalence.