Resolvendo equações

Math in Music

Now that we have an understanding of what an exponential function is, we're going to look at a specific use of exponential equations in music.

If you've ever listened to a piano, you may have noticed that each note going up the entire piano sounds an equal distance apart. This is due to our modern tuning system known as equal temperament. By making each note on a piano sound an equal distance apart, we can play music in any key on a piano without getting any clashes. The downside to this is that many of the intervals are not perfectly in tune with the ratios used to tune some intervals.

When the piano itself is tuned you first pick a starting note, normally $A=440\text{ Hz}$. From there every octave of $A$ is tuned. Each octave will multiply the pitch by a power of $2$ (written as $2^x$) and this will give you a baseline to work with. From there, you will need to place the remaining pitches at $12$ equal sounding spaces between the octaves, due to the $12$ notes in each octave.

The placement of these pitches are not as easy as dividing into 12 equal spaces, as human hearing of pitches doesn't follow a linear pattern. The distance between $220\text{ Hz}$ and $440\text{ Hz}$ sounds the same as the distance between $440\text{ Hz}$ and $880\text{ Hz}$. Humans have logarithmic based hearing in order to counteract the exponential nature of pitches. This means that to find $12$ equal sounding notes in between each octave, the tuner must use the expression $2^\frac{x}{12}$.

If we want to find the exact pitch in Hz for any note on a regular piano, we can use the following expression: $$440 \cdot 2^\frac{x}{12}$$ In this expression, $x$ is the number of half steps away from the original note, $A=440\text{ Hz}$, the note we are trying to find is. Using the picture above, $A$# is one half step away from $A$, so we would use $440\cdot 2^\frac{1}{12}$ to find $A$#'s pitch. Similarly, $D$ is $5$ half steps above $A$, so we would use $440 \cdot 2^\frac{5}{12}$ to find the pitch for $D$.

Let's try out a problem to see if we understand how the equation works.

Here all we have to do is make sure we put the correct values into the equation. We know that this is a standard piano, so the entire tuning is based around $440\text{ Hz}$. Using the picture above, we can see that $F$ is $8$ half steps above $A$, so we can plug that in for $x$. $$\begin{align} \text{Pitch of F}&= 440 \cdot 2^\frac{8}{12}\\ \text{Pitch of F}&= 440\cdot .667\\ \text{Pitch of F}&= 698.46 \end{align}$$

Depending on the information given to us, we can solve this equation to find different pieces of information. Let's define this equation in general terms.

As long as you have any two pieces of information, you can solve for the remaining information.

Let's try a practice question.

First we're gonna set up our problem. Matt is playing on a standard piano, so his tuning note is at $440\text{ Hz}$.

$$\begin{align} 493.88&= 440 \cdot 2^\frac{x}{12}\\ \end{align}$$

Now we can solve this like we would any other exponential equation. We're going to begin by dividing both sides by $440$ to begin to isolate our variable.

$$\begin{align} \frac{493.88}{\color{#8925AE}{440}}&= \frac{440 \cdot 2^\frac{x}{12}}{\color{#8925AE}{440}}\\ 1.12&=2^\frac{x}{12} \end{align}$$

Just a quick note, through this problem we are always going to be rounding to the hundredths place. This may take our answer a little bit away from the true values, but because it's music we can just pretend its due to a poorly tuned instrument.

Now we're going to take the log of both sides and rearrange our problem using exponent and logarithm laws in order to solve. Remember that $x=y\Longleftrightarrow\log_ax=\log_ay$.

$$\begin{align} 1.12&=2^\frac{x}{12}\\ \log1.12 &= \log2^\frac{x}{12}\\ \log1.12 &= \frac{x}{12}\log2 \end{align}$$

From here all we need to do is isolate $x$.

$$\begin{align} \log1.12 &= \frac{x}{12}\log2\\ \frac{\log1.12}{\log2} &= \frac{x}{12}\\ 12 \cdot \frac{\log1.12}{\log2} &= x\\ x &\approx 1.96 \end{align}$$

Now you may be thinking "but Jonathan this isn't one of the options you gave me!!!!". Everything is okay, we just need to think about this from the context of the problem. In musical terms, a note will never be a percentage of a full half step away from a different note. $493.88\text{ Hz}$ is equivalent to a $B$, which is $2$ half steps away from the tuning note of $A$. Our final calculated answer was $1.96$, but here we can round that up to $2$ to make sense in the context of the problem.

(Video) Solving Exponential Equations

by PatrickJMT

Here's a video by PatrickJMT showing you some different approaches to solving exponential equations. The main idea to solve these is to take the natural log of each side. This brings the exponential part out front so you can isolate it.

(Video) Exponential Equations

by mathman1024

Here's a video by mathman1024 where he solves exponential equations.

$\large{\text{Summary}}$

The key property when solving these problems is: $$\text{If }a^m=a^n\\ \text{then }m=n$$ This means that if the bases are the same then the exponents are the same.

The first problem he looks at is: $3^{x+2}=27$. At first glance you might think they don't have the same bases but we can actually rewrite the $27$. $$\begin{align} &3^{x+2}=27\\ &3^{x+2}=\color{#8925AE}{3^3} \end{align}$$ So since the bases are the same then the exponents have to be the same. Which means: $$\begin{align} &x+2=3\\ &x+2\color{#8925AE}{-2}=3\color{#8925AE}{-2}\\ &x=1 \end{align}$$

Try the next problem on your own!

Now that we've learned about exponential growth and exponential decay, we can learn how to solve exponential equations. We'll then move into asymptotes and graphing exponential equations.

Solving Exponential Equations with Like Bases

When you're solving exponential equations with like bases, the goal is to rewrite both sides of the equation so they have the same base. An exponential equation has the form:

$$\text{base}^{\text{exponent}} = \text{base}^{\text{exponent}}$$

This means both sides of the equation are written as a power with the same base.

Once the bases math, you can use a property of exponents that says that:

If $a^x = a^y$, then $x=y$, as long as $a$ is a positive real number and not equal to $1$.

This property allows you to set the exponents equal and then solve the resulting equation. That step is called solving the equation.

Example

For example, if you're given: $3^{x+2} = 3^5$

Since both sides have a power with base $3$, you can now say: $x+2=5$

Then you can solve for the unknown: $x=3$

This works because of the rules of exponents and because equations like this have a solution when the bases are the same. Always remember that this only works when the sides of the equation have the same base.

If the bases are different, you may need to rewrite both sides using a common base before applying the rule. Once the equation is in the right form, solving it becomes as simple as solving a normal equation with variables.