# Notes on a^x and loga(x)

Here is an approach to the exponential and logarithmic
functions which avoids any

use of integral calculus. We use without proof the existence of certain limits
and assume

that certain functions on the rational numbers can be extended to continuous
functions on

the reals. All of this can be justified, but we do not do so here.

Let a be a positive real number . We want to define a^{x}. For n a
positive integer, a^{n}

is a multiplied by itself n times. Similarly, a^{-n} is a^{-1} multiplied by itself n
times. Every

positive number has a unique positive m- th root , so we can define (for m a
positive integer)

Having defined a^{x} for x a rational number, we define a^{x}
for all real x by choosing a sequence

of rationals converging to x, etc. This process leads to a well defined function
a^{x} which is

a continuous function from the whole real line to the positive reals.

**Proposition 1. ** and
. Moreover, a^{0} = 1 and a^{1} = a.

**Proof.** The first two assertions follow by first proving them for x and y
rational and then

using continuity. To show a^{0}= 1, set x and y both equal to 1 in the second
identity. That

a^{1} = a follows from the definition.

**
Lemma 2. **Assume a > 1. Then a

^{x}is a strictly increasing function.

**Proof.**Suppose x < y and that x and y are rational. By adding a positive integer to both

sides, we can assume that x and y are positive. Write both over a common denominator

N. Thus, x = m/N and y = n/N. Since x < y, we have m < n. Now, a > 1 implies

. Thus,

Having established the result for rational numbers, the
general result follows by taking

limits.

**Definition. **For a > 0, the limit

exists. It is called ln(a), the natural logarithm of a.

**Proposition 3.** The derivative of a^{x} exists, and we have

**Proof. **This follows easily from the definitions

**Remark:** note the above proposition is similar to how we differentiated trigonometric

functions. To find the derivative of sin(x) or cos(x) at any point, we needed to
compute

two limits ,and
, the derivative of sine and cosine at any

point followed by the angle addition formulas . Here, the analogue is
.

**Corollary**. If a > 1 then ln(a) > 0. For a = 1, ln(1) = 0. If 0 < a < 1, then
ln(a) < 0.

**Proof.** The second assertion is clear since 1^{x} = 1 for all x and so its
derivative is zero.

To prove the first assertion, note that if ln(a) < 0 the by the Proposition, a^{x}
is

decreasing. This contradicts Lemma 2. Thus, ln(a) ≥ 0 if a > 1. However, it
can't be 0,

since then the derivative of a^{x} would be identically zero by Proposition 3, and
a^{x} would be

a constant. This again contradicts Lemma 2. Thus, ln(a) > 0. Finally, if a < 1,
we have

a^{x} = ((a^{-1})^{x})^{-1} is decreasing, so, by repeating the reasoning above, we find ln(a) < 0.

**Remark: **We do not consider ln(a) for a ≤ 0.

**
Proposition 4. **Assume a, b > 0. Then ln(ab) = ln(a) + ln(b).

**Proof.**By definition,

As

we have

Therefore

As h → 0, note b^{h} → 1, and we therefore find that

log(ab) = ln(a) + ln(b).

Then, ln(ab) = ln(a) + ln(b) .

**Proof. **We have already proved the first assertion, but here is a second proof
using

Proposition 4.

ln(1) = ln(1 · 1) = ln(1) + ln(1),

which implies ln(1) = 0.

Now, 0 = ln(1) = ln(a · a^{-1}) = ln(a) + ln(a^{-1}). The second assertion follows.

**Lemma 5.** ln(x) is a strictly increasing function.

**Proof.** Suppose 0 < x < y. Then, 1 < y/x and ln(y/x) > 0 by the Corollary to

Proposition 3. Thus ln(y) - ln(x) > 0, or ln(x) < ln(y), which proves the
assertion.

**Proposition 6.** For all positive a, we have ln(a^{x}) = x
ln(a) .

**Proof.** Let f(w) = ln(a^{x}). Using Proposition 4, one easily
checks that

. By simple algebra one deduces that
for all
rational

numbers r. Since f(w) is continuous, the Proposition follows for all real
numbers x.

**Corollary.** As x → ∞, ln(x) → ∞. Also, as x → 0, ln(x) → -∞.

**Proof. **Since ln(x) is increasing we only have to prove that it takes on
arbitrarily large

values as x gets bigger and bigger. Consider the sequence
We have

ln(2^{n}) = n ln(2) → ∞ as n → ∞.

Similarly, on the sequence we have

as n → ∞ .

This completes the proof.

**Definition.** The following limit exists

We call this limit e, Euler's constant, e is approximately
2.71828. Note that another

expression for e is given by

**Proposition 7.** The function ln(x) is differentiable. We
have

**Proof.** As , using
Proposition 4 we have

Therefore we have

Using the definition of the derivative, we see that

Letting , as h → 0 we see h' → 0. By Proposition 6,

and by the above definition this limit is just ln(e).
Combining the pieces gives the derivative

of ln(x) is , as claimed.

**Proposition 8**. ln(e) = 1 .

**Proof.** By Proposition 6, we have

Differentiate both sides using what we have proven and, of course, the chain
rule.

Remember that e is a constant, so ln(e) is just a number { it has no x
dependence. Thus,

the derivative with respect to x of ln(e) is zero , and the derivative with
respect to x of

x ln(e) is thus ln(e).

We use the chain rule to differentiate ln(e^{x}). Let f(x) = ln(x) and let g(x) =
e^{x}.

Then ln(e^{x}) = f(g(x)), so by the chain rule its derivative is f'(g(x))
· g'(x).
We get the

derivative of f from Proposition 7, and the derivative of g from Proposition 3.
Substituting

gives

We have shown that this derivative is also equal to ln(e),
therefore we find that [ln(e)]^{2} =

ln(e). Since ln(e) ≠ 0, we must have ln(e) = 1.

**Corollary.**

**Definition. **We define the logarithm function to the base a
by the following formula

This function has all the properties one would expect. We
list them. The proofs are

very easy and are left to the reader.

6.

7. If a > 1, then is strictly increasing and its
graph is everywhere concave down.

It goes to ∞ as x → ∞ and to minus ∞ as x → 0.

Finally, we note that , so that we have

and .

What this means is that if then x = a^{y}. In other words,
is
the number

of powers of a we need to get x.

For example, consider : we raise a to the number of powers of a we need
to

get x, thus .

Finally, note that in general, the logarithm of
a sum

is** not** the sum of the logarithms.

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