## Exponential Graphs

Exponential graphs are those of the form $y=a^x$ for positive a. You can sketch the graph of $y=a^x$ for positive a by considering the y coordinates that correspond to various x values. Graphs of this form will always cross the y-axis at 1 since $a^0=1$ for any a.

The following table shows coordinates for the graph $y=2^x$ for x taking integer values between -3 and 3:

x -3 -2 -1 0 1 2 3
y 8 4 2 1 0.5 0.25 0.125 It follows that the graph of $y=a^x$ for $a\textgreater1$ will have a shape like $y=2^x$.

The graph of $y=a^x$ for $0\textless a\textless 1$ will have a shape like the graph above but will be reflected in the y-axis. This is because when you multiply a number less than 1 by itself, it becomes smaller.

The following table shows coordinates for the graph $y=0.5^x$ for x taking integer values between -3 and 3:

x -3 -2 -1 0 1 2 3
y 0.125 0.25 0.5 1 2 4 8 For any number $0\textless a\textless 1$, the graph will have the same shape as $y=0.5^x$.

For a=1, the graph of $y=a^x$ is the horizontal line $y=1$. This is because you are calculating 1 to any power, which is always 1. For negative a, fractional powers become an issue and complex numbers need to be considered. The diagram shows the graph of $y=e^{x}$ where e, sometimes known as Euler’s number, is given by e=2.718281828459… Since e is positive and greater than 1, it looks very similar to the first graph above.

The number e is special because everywhere on this graph, the gradient is the same as the y-coordinate. See differentiating e to the x.

## Logarithmic vesus Exponential Graphs As well as exponential graphs, there are logarithmic graphs. $\log_a(x)$ is considered to be the inverse of $a^x$ – see more on logs. It follows that $\log_a(x)$ (blue solid line) is the inverse of $a^x$ (red dotted line) and so their graphs are reflections of each other in the line y=x (green dotted line). Since $\log_a(x)$ and $a^x$ are mathematical inverses we have that $\log_a\left(a^x\right)=a^{\log_a(x)}=x$.
Specifically, the natural logarithm is the logarithm that corresponds to e. That is, given an equation of the form $y=e^{x}$, it can be said that $x=\log_e(y)$. Since e is a special number, log to the base e has its own name. That is, it is the natural logarithm and often called ln so $x=\log_e(y)$ is more often written as $x=\ln(y)$. $\ln(x)$ and $e^x$ are mathematical inverses and we have that $\ln\left(e^x\right)=e^{\ln(x)}=x$. Notice that $log_a$ for any positive a (including ln) cannot be evaluated for negative x – see more on logs.

YouTubeGraphing Logarithmic Functions

### Estimating Parameters for $y=ax^n$

Consider the equation $y=ax^n$. Note that, according to BIDMAS, this is x to the power of n, then multiplied by a. This is a stretch to a standard polynomial curve – see Curve Sketching. Given this relationship and a dataset that approximately fits it, it is possible to estimate the parameters a and n. First consider what happens when logging both sides:  $\begin{array}{c}\log(y)=\log(ax^n)\\\Longrightarrow \log(y)=\log(a)+\log(x^n)\\\Longrightarrow \log(y)=\log(a)+n\log(x)\end{array}$

Note that the bases are missing this is true for any base (provided the same base is used for both). In the same way that you can plot y against x, it is possible to plot log(y) against log(x). Recall that, in the equation y=mx+c, m is the gradient and c is the y-intercept. In addition, we can write log(y) as nlog(x)+log(a) and so, in the plot of log(y) against log(x), n is the gradient and log(a) is the y-intercept.

#### Example

The following table follows the relationship $y=ax^n$ where the y values are given to one decimal place. By plotting $\log(y)$ against $\log(x)$, this allows us to estimate the parameters a and n to 1 decimal place.

x 2 3 4 5
y 26.4 104.7 278.6 594.9

### Estimating Parameters for $y=kb^x$

Now consider the equation $y=kb^x$. Like the above, this is an exponential curve, provided b is positive. Similarly to before, given a dataset or similar, we could estimate the parameters b and k. Taking logs:  $\begin{array}{l}\log(y)=\log(kb^x)\\\Longrightarrow \log(y)=\log(k)+\log(b^x)\\\Longrightarrow \log(y)=\log(k)+x\log(b)\end{array}$

It follows that log(y) can be written as log(b)x+log(k) and so, this time, in the plot of log(y) against x, log(b) is the gradient and log(k) is the y-intercept.

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