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Other Morphs

Using symbol motion and fusion will let you solve many equations intuitively. There are also situations where you need something else. Take this equation:

$$(x-2)^{2}=4$$

The easiest solution path involves taking the square root of each side^5.1:

$$\sqrt{(x-2)^{2}}=\sqrt{4}$$

$$x-2=2$$

There is not a way to express square roots using symbol motion, fusion or fission. We need another kind of morph. We are now getting into an area where the technique is less structured than what you've seen so far. The benefit is that such morphs are more flexible, and even somewhat adaptable.

Let's make this concrete by showing how morphs are used to solve square roots. The process goes something like this:

1. Set up the equation so that you can apply the square root to both sides.
2. Add symbols to each side that describe that operation.
3. Simplify each side, one at a time, using all the techniques at your disposal - symbol motion, fusion, or other morphs.

Here is an example of how that works. As always, first visualize the equation.

The first step in applying the morph is to visualize a square root around each side:

Next, in the image, we substitute the changed expression, on one side at a time. So starting with the image just above, morph on the left side first,

Then we do the other side.

Instead of a graphical substitution, you can also do an animation. So with the change from $\sqrt{(x-2)^{2}}$ to x - 2, you can imagine the $\sqrt{(\,\,\,\,)^{2}}$ part fading out. The animation from $\sqrt{4}$ to 2 will be more fanciful; most morphs will be like that.

Another use of morphs is when working with equations that use exponents, and which you would solve using logarithms. The process is quite similar to what we just did with the square root. For example, say you have this equation,

$$A(t)=A_{0}(1+e^{a+bt})$$

Here $A_{0}$, a, and b are constants, t is the independent variable, and A = A(t) is the dependent variable. This one is very similar to equations like $A(t)=A_{0}e^{bt}$, which you have probably seen at some point in a textbook (or will see). You want to invert this equation, so that instead of $A(t)=something$ you have $t(A)=something$. If it was the textbook form, you may actually remember the inverted form, but since it is different you will have to invert it yourself. Let's do that.

We start by visualizing the equation, as always.

Remember that you can use windowing to make visualizing it easier (page ). Next, using symbol motion, rearrange the equation so that the exponential is alone on one side:

Now we will use the morph. Apply the natural log to each side.

Remember that there is a property of logarithms, that $\log a^{b}=b\log a$. Because of this, you can use symbol motion to bring the a + bt in the exponent outside of the log:

(This is a use of symbol motion that you did not see before. You will occasionally find interesting new places you can apply it like this.) The next step is to deal with the $\ln e$, which you do by chunking it,

then morphing into nothing.

From there, you can use symbol motion to complete the inversion:

Remember what was said in the last section, about crossing the pond using stepping stones? And that with experience, you will be able to take larger and larger 'jumps' in your algebra calculations? This scenario is one place where this can happen. Above, our process changed $\ln e^{a+bt}$ into $(a+bt)\ln e$, then to a + bt. After you have done this a while, you will go from $\ln e^{a+bt}$ to a + bt in one step inside your mind. You will chunk the $\ln e^{a+bt}$, and morph that chunk directly into a + bt. This kind of progression can apply to all morphs (actually, any mental math you do at all). As you get more experience, you will discover you can do bigger substitutions and changes at once by using bigger or different morphs.

Morphs can be arbitrarily general. In other words, there are other kinds of morphs than what are shown in this chapter. Most of the morphs used in doing algebra work are introduced in this book. Other morphs and similar techniques can be used in other areas of math, such as calculus.^5.2

Footnotes

... side^5.1

More properly, we'd end up with $x-2=\pm2$. We'll just work with the positive root here, since we haven't learned how to deal with equations with multiple solutions yet. (That comes in chapter , ``Plurality''.)

...^5.2

The book Inner Math: How To Do Algebra, Calculus, and more In Your Head, which is a sequel to this one, will show how to do this.