iGCSE (2021 Edition)

# 11.12 Defining e and the natural logarithm

Lesson

### Euler's number "$e$e"

In calculus, when we differentiate a function this gives us its gradient function. We can then use this function to find the gradient of a tangent at any point. So far, we haven't learnt a process for differentiating an exponential function. This is algebraically quite different to the functions we have seen using powers of $x$x, as opposed to a number raised to the power of $x$x

Interestingly though, our knowledge of sketching derivative functions leads us to the observation that the derivative functions of exponentials look quite similar to their original functions. Below is a graph of the exponential function $y=2^x$y=2x, along with the graph of its derivative (the dashed line). Pretty close, don't you think?

In the applet below, drag the slider to change the base of the exponential. Try to find an approximate value for the base at which the gradient function lines up and matches with the original function:

 Created with Geogebra

Both curves overlap at around $2.72$2.72:

This tells us that there is a value of the base $a$a, which is approximately $2.72$2.72, for which the derivative of the function is equal to itself. This particular base is known as Euler's number, and is given the symbol $e$e. It is named after the mathematician Leonhard Euler, and although you may not have seen it before, this irrational number is as famous as $\pi$π.

We call the function $y=e^x$y=ex "the exponential function" to distinguish it from all other exponential functions.

If we want to differentiate an exponential function, we require that it is written in base $e$e. Next year, we will see how we can differentiate exponential functions with other bases. We will do this by first changing them to base $e$e. But for now, let's just get used to this new, rule for differentiation.

The derivative of $e^x$ex

$\frac{d}{dx}\left(e^x\right)=e^x$ddx(ex)=ex

If $y=e^x$y=ex, then we can also express the derivative as $\frac{dy}{dx}=y$dydx=y

This result tells us that the gradient of an exponential function at a point is always equal to its height above the $x$x-axis at that point.

### Other ways to approximate $e$e

There are many ways to calculate $e$e. Obviously, the button on our calculator is a great way to approximate $e$e. You can find the button on the calculator to approximate values with $e$e near the "ln" button.

Here are two other ways:

• Consider the value of $\left(1+\frac{1}{n}\right)^n$(1+1n)n as n gets bigger. In fact, use your calculator to evaluate $\left(1+\frac{1}{10000}\right)^{10000}$(1+110000)10000. What do you get?
• The value of $e$e is also found by the infinite sum: $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\ldots$10!+11!+12!+13!+14!+15!+ (the symbol "!" means factorial). If you type the first six terms of the sum into the calculator it will give you $2.7166666667$2.7166666667. The more terms you add, the better the approximation.

### The graph of the exponential function

The usual methods of shifting and reflecting graphs can be applied to $y=e^x$y=ex. The graph of $y=e^x$y=ex is one of the most important graphs in this course and we need to be very familiar with its shape and properties.

Properties of the exponential function:

• The domain is all real $x$x, the range is $y>0$y>0
• There are no $x$x-intercepts as the curve is always above the $x$x-axis
• The $x$x-axis ($y=0$y=0) is a horizontal asymptote to the curve
• The curve has a gradient $1$1 at it's $y$y-intercept $\left(0,1\right)$(0,1). (We can see this by substituting $x=0$x=0 into$y'=e^x$y=ex)
• The curve is always increasing at an increasing rate and is always concave up.

#### Worked example

##### Example 1

Find the gradient of the tangent to the curve $y=e^x$y=ex at the points:

(a) $\left(0,1\right)$(0,1)

Do: The gradient function is given by $y'=e^x$y=ex

Substituting$x=0$x=0 into the gradient function gives:

$y'=e^0$y=e0

$=1$=1

Therefore the gradient is 1 at the point $\left(0,1\right)$(0,1)

(b) $(-1,1/e)$(1,1/e)

Do: Substituting $x=-1$x=1 into the gradient function gives:

$y'=e^{-1}$y=e1

$=\frac{1}{e}$=1e

#### Practice question

##### QUESTION 1

$f\left(x\right)=e^x$f(x)=ex and its tangent line at $x=0$x=0 are graphed on the coordinate axes.

1. Determine the gradient to the curve at $x=0$x=0.

2. Evaluate $f\left(0\right)$f(0).

3. Which of the following is true?

$f\left(0\right)\ne f'\left(0\right)$f(0)f(0)

A

$f\left(0\right)=f'\left(0\right)$f(0)=f(0)

B

$f\left(0\right)\ne f'\left(0\right)$f(0)f(0)

A

$f\left(0\right)=f'\left(0\right)$f(0)=f(0)

B

## Natural logarithms $\log_ex$loge​x or $\ln x$lnx

Natural logarithms are logarithms to the base $e.$e.We call this the "logarithmic function" to distinguish it from other logarithmic functions with bases other than $e$e.

When we rearrange $y=e^x$y=ex into logarithmic form we get the natural logarithmic function $y=\log_ex$y=logex, which is also written as "$\ln x$lnx" (short for "natural logarithm").

The "$\ln$ln" button on the calculator can help us evaluate logarithmic functions with base $e$e. The log laws that we previously studied also applies to natural logarithms to help us simplify log expressions and equations.

#### Practice questions

##### Question 2

Find the value of $\ln94$ln94 correct to four decimal places.

##### Question 3

Find the value of $\ln\left(18\times35\right)$ln(18×35) correct to four decimal places.

##### Question 4

Use the properties of logarithms to express $\ln\sqrt[3]{y}$ln3y without any powers or surds.

### The graph of $y=\log_ex$y=loge​x

The exponential and the logarithmic functions are inverse functions which means that their graph will be a reflection of each other across the line $y=x$y=x (their $x$x and $y$y values are swapped).

Because they are reflections, the properties of the natural logarithm graph will correspond with the properties of $y=e^x$y=ex:

• The domain is $x>0$x>0, the range is all real $y$y
• The $y$y-axis ($x=0$x=0) is a vertical asymptote to the curve
• The curve has gradient $1$1 at its $x$x-intercept $\left(1,0\right)$(1,0)
• The curve is always decreasing and concave down

### Combining exponential and logarithmic functions

Recall from the definition of logarithms that if $y=\ln x=\log_ex$y=lnx=logex then $x=e^y$x=ey

So if we substitute $x=e^y$x=ey into $y=\ln x$y=lnx, we see that $y=\ln e^y$y=lney. In other words, raising $e$e to the power $y$y, and then taking logs on that answer restores the original $y$y

If we substitute $y=\ln x$y=lnx into $x=e^y$x=ey, we can also show that $x=e^{\ln x}$x=elnx. Again taking the log of $x$x, and then use that answer as the exponent that $e$e is raised to, simply restores the original $x$x.

The Logarithmic and Exponential Functions are inverse functions

$\log_ee^x=x$logeex=x and $e^{\log_ex}=x$elogex=x

Using the above argument, the expression $e^{\ln5.4}$eln5.4 is simply $5.4$5.4. Also, the expression $\ln e^{\sqrt{5}}$lne5 is simply $\sqrt{5}$5.

### Outcomes

#### 0606C14.3C

Use the derivatives of the standard functions e^x, ln x, together with constant multiples, sums and composite functions.