Intro to exponential functions (y=a^x, a^-x)

It is not too difficult to make sense of expressions with an exponent when the exponent is an integer. For example, we interpret $2^3$23 to mean $2\times2\times2$2×2×2 and $2^4$24 to mean $2\times2\times2\times2$2×2×2×2, and so on.

But it is not so easy to give a meaning to an expression like $2^{3.5}$23.5 or, even worse, $2^{\pi}$2π.

If these expressions are to mean anything at all, we would have to agree that they are real numbers and that $2^3<2^{\pi}<2^{3.5}<2^4$23<2π<23.5<24. More generally, $2^a<2^b$2a<2b whenever $aa<b.

 

We will want the rules for multiplying and dividing numbers with integer powers to continue to be valid when the exponents are not necessarily integers but could be any number.

Recall that

$a^m\times a^n=a^{m+n}$am×an=am+n
$a^m\div a^n=a^{m-n}$am÷​an=am−n
$\left(a^m\right)^n=a^{mn}$(am)n=amn
$a^{-1}=\frac{1}{a}$a−1=1a​
$a^0=1$a0=1 for all $a\ne0$a≠0
$a^{\frac{1}{n}}=\sqrt[n]{a}$a1n​=ⁿ√a. In particular, $a^{\frac{1}{2}}=\sqrt{a}$a12​=√a.

 

If we can convince ourselves that expressions with real number exponents do make sense, then we can construct functions which we will call exponential functions. These are of two main kinds. They have the form

$y=a^x$y=ax or  $y=a^{-x}$y=a−x
 

We understand $x$x to be the domain variable where $-\infty−∞<x<∞ and $a$a to be a constant which we call the base. Complications arise if we allow the base to be a negative number, and with zero as the base, the function would have the value zero for every non-zero $x$x. So, we will require that any base we choose for an exponential function will be a positive number.

(We will see in the examples below that we could also restrict the base to be greater than $1$1.)

 

We have seen that $a^0=1$a0=1 for all bases $a$a. So, the graph of every exponential function must pass through the point $(0,1)$(0,1) on the vertical axis.

Also, if $u$u and $v$v are any two numbers in the domain and $uu<v, then $a^uau<av. So, the function $a^x$ax must be an increasing function.

On the other hand, if $uu<v, then $\frac{1}{a^u}>\frac{1}{a^v}$1au​>1av​. So, the function $a^{-x}$a−x must be a decreasing function.

 

As might be expected, exponential functions are steeper for larger bases. For example, $10^x$10x is always going to be a larger number than $2^x$2x when $x$x is positive and the comparison is reversed when $x$x is negative.

 

To see what the value must be of the expression $2^{3.5}$23.5 which we introduced at the beginning of this discussion, we could think of it as $\left(2^7\right)^{\frac{1}{2}}$(27)12​. This is $\sqrt{128}\approx11.31$√128≈11.31. Note that $11.31$11.31 is less than the number mid-way between $2^3$23 and $2^4$24. The graph must be getting steeper with increasing $x$x. In fact, this increasing steepness is what is meant by the everyday expression exponential growth.

We could perform a similar trick with other bases and with other rational exponents to see how the exponential functions we are constructing must behave.

This will still not tell use what value we should assign to an expression like $2^{\pi}$2π that has an irrational exponent. But if we pick two rational exponents close to $\pi$π, one on either side, then we can narrow down a value for $2^{\pi}$2π as accurately as we wish.

 

 

Example 1

Suppose we have chosen the number $\frac{1}{3}$13​ to be the base of an exponential function which we write as $\left(\frac{1}{3}\right)^x$(13​)x. Show that this can be written more simply without a fractional base.

According to the rules for manipulating exponential expressions, we have

$\left(\frac{1}{3}\right)^x$(13​)x	$=$=	$\frac{1}{3^x}$13x​
	$=$=	$\left(3^x\right)^{-1}$(3x)−1
	$=$=	$3^{x\times(-1)}$3x×(−1)
	$=$=	$3^{-x}$3−x

Thus, an exponential function with a fractional base is equivalent to a negative exponential function with a base greater than $1$1.

Note that if the base is $1$1, then the function has a constant value of $1$1 for every $x$x.

 

Example 2

Assign an approximate value to the exponential function $y=1.2^x$y=1.2x when $x=\sqrt{3}$x=√3.

The exponent $\sqrt{3}$√3 is an irrational number. Consider the following sequence of rational numbers:

$2,\frac{5}{3},\frac{7}{4},\frac{19}{11},\frac{26}{15},\frac{17309}{9999},...$2,53​,74​,1911​,2615​,173099999​,...

The numbers oscillate around $\sqrt{3}$√3 and get ever closer to it. For example, $\sqrt{3}$√3 is somewhere between the rational numbers $\frac{19}{11}$1911​ and $\frac{7}{4}$74​. We could conceivably, with a great deal of work, compute by hand the roots $\sqrt[11]{1.2^{19}}$¹¹√1.219 and $\sqrt[4]{1.2^7}$⁴√1.27 to, say, four decimal places.

We can do this with a calculator, without for the moments asking how the machine does it. We evaluate $1.2^{\frac{19}{11}}$1.21911​ and $1.2^{\frac{7}{4}}$1.274​. and obtain respectively $1.370$1.370 and $1.376$1.376.  

The number we seek must be somewhere in-between. Again, mysteriously, the calculator gives $1.2^{\sqrt{3}}=1.371...$1.2√3=1.371...

We could go further along the sequence to obtain ever smaller intervals presumed to contain the number $1.2^{\sqrt{3}}$1.2√3. We do not need to perform the onerous calculations for this in order to be convinced that such a number must exist. Then, having reached this conclusion, we can go on to accept that the function given by $1.2^x$1.2x makes sense for all real number values of $x$x.

 

Worked Examples

Question 1

Question 2

Question 3