# Math Insight

### Redundant parameters in the exponential function

One way to add parameters to the exponential function is to let the parameter $b$ be the base and the parameter $k$ multiply the exponent, forming the function $$f(x)=b^{kx}.$$ Since the function $f$ has two parameters, $b$ and $k$, it can be represented by a function machine with two dials representing the parameters. #### Redundant

But do these two dials do different things? Does changing the $b$ dial really do something different than changing the $k$ dial? Obviously, they show up in the function in different ways, so a given change to both parameters doesn't do exactly the same thing. Doubling $b$ will have a different effect than doubling $k$. But maybe a different change in $b$ will have the same effect as doubling $k$.

In fact, the two parameters $b$ and $k$ do have the same effect. It turns out that you can square the value of $b$ to give the same effect as doubling the value of $k$. You can explore this fact with the following applet. If you click the “fix function” checkbox, you can see how the effect of any change in one parameter can be undone by a corresponding change in the other parameter.

The exponential function. The exponential function $f(x)=b^{kx}$ for base $b >0$ and constant $k$ is plotted in green. You can change the parameters $b$ and $k$ by typing new values in the corresponding boxes. It turns out the parameters $b$ and $k$ can change the function $f$ in the same way, so you really only need to change one of them to see all the different functions $f$. To see how they do the same thing, you can click the “fix function” checkbox, which will fix the function $f(x)$. When that box is checked, if you change the parameters $b$ or $k$, the other parameter will change in a way to leave the function $f(x)$ unchanged. For the function $f(x)=b^{kx}$, the value $f(0)=1$ for all parameters. To change the value of $f(0)$, you can allow scaling of the function by clicking the corresponding checkbox. Then, the function changes to $f(x)=c b^{kx}$ with an additional parameter $c$ that scales (multiplies) the whole function so that $f(0)=c$. You can change the value of $c$ by dragging the red point. You can change range of the $x$ and $y$-axes buttons labeled $x+$, $x-$, $y+$, and $y-$. Since $f(x)$ is always non-negative, only the positive $y$-axis is shown.

To see why this is true, we can do a simple calculation. Understanding this calculation requires that you are comfortable with the basic rules of exponents and logarithms.

In the original form of the function, $f(x)=b^{kx}$, it's hard to see any correspondence between $b$ and $k$ because they have very different relationships with the independent variable $x$. The parameter $k$ multiplies $x$, but the parameter $b$ is the base for the exponent involving $x$. The first step in fixing this problem is to take the logarithm of both sides of the equation $f(x)=b^{kx}$: $$\log f(x) = \log b^{kx}.$$ (For this calculation, it doesn't matter which base we use for the logarithm.)

At first glance, this manipulation might not seem like an improvement. But, the properties of the logarithm allow us to make an important simplification. We can take the exponent $kx$ down from $b$ and move it to multiply $\log b$. \begin{align*} \log f(x) &= \log b^{kx}\\ &=kx \log b\\ &= x(k \log b) \end{align*} The final expression $x(k \log b)$ emphasizes that both $k$ and $\log b$ have the same relationship with the independent variable $x$. It's really not the separate values of $k$ and $b$ that matter in determining the function. As long as I know the value of the combination $k \log b$, I don't care about the individual values of $k$ and $b$. The combination is enough to determine $\log f(x)$, and hence the original function $f(x)$ itself.

When you check the “fix function” checkbox, the applet uses the expression $k \log b$ to calculate how to change one parameter to compensate for your change in the other. If you change $b$, it just finds the value of $k$ so that $k \log b$ is the same value as before you made the change.

For example, imagine you start with $k=2$ and $b=2$, which means $k \log b =2 \log 2$. Now, you change change $k$ to $k=1$. To keep the function fixed, we need to find the value of $b$ so that $1 \log b = 2 \log 2$. Given the properties of the logarithm, we can write this condition as $$\log b = 2\log 2 = \log 2^2.$$ In other words, we need $b=2^2=4$.

In general, if you cut $k$ in half, you need to square $b$ to cancel its effect. Or, thinking of it another way, doubling $k$ is the same thing as squaring $b$.

We can conclude that we made our lives more complicated than we needed to by introducing both the parameter $b$ and the parameter $k$. We can get rid of one of them without losing anything. You might think the simplest thing to do would be to get rid of $k$ and write $f$ as $$f(x) = b^x.$$ We do this sometimes. But, for reasons that must remain mysterious until you've learned some calculus, we will commonly fix the exponential base to the irrational number $$e= 2.718281828459045 \ldots .$$ Since we fix $b=e$, we need to keep the parameter $k$ to change the shape of the function $f$, writing $f$ in the form $$f(x) = e^{kx}$$ with the single parameter $k$. (The applet understands the value of $e$, so you can type $e$ in the box for $b$.)

#### Non-redundant

We can add a second parameter to $f$ that is not redundant with $b$ or $k$. Notice that for any value of $b$ or $k$, $f(0)=b^{k0}=1$. To allow $f(0)$ to be a different value, we can introducing a scaling parameter $c$, defining $$f(x)=c b^{kx}.$$ Then $f(0)=c$. Since a change in $c$ cannot be reproduced by changing either $k$ or $b$, the parameter $c$ is not redundant and allows the function $f$ to have a different behavior. You can add in the parameter $c$ by checking the “allow scaling” checkbox in the above applet.

If we fix $b=e$, then the function $$f(x)=ce^{kx}$$ has two parameters that do very different things. The parameter $c$ scales the function and determines the value of $f(0)=c$. The parameter $k$ determines how quickly the function increases or decreases.